The cursor inside stored procedure not working as expected - sql-server

I have an sql stored procedure, but when I execute it it is not returning the desired out put. Can someone please tell me what I'm doign wrong here...
I put some print statements in there to debug.. the problem seems to be that the cursor is not fetching any results. the result of ##fetch_status is -1
Here is my stored procedure in question:
ALTER PROCEDURE [dbo].[spComplaintsEmailAlert_autoFillForm] (#counterVal int,
#RESULT varchar(50) OUTPUT,
#RESULTDESC varchar(max) OUTPUT)
AS
DECLARE #dateEmailSend varchar(50),
#counter int,
#region varchar(50),
#ola_company varchar(50),
#obj_identified_as varchar(50),
#product_involved varchar(50),
#ola_customer varchar(50),
#product_ut varchar(50);
DECLARE otherCounterVals CURSOR FOR
SELECT
A.CounterVal
FROM [cmpEmailAlerted] A,
[cmpTblProductComplaints] B
WHERE A.CounterVal = B.Counter
AND CONVERT(varchar(10), A.dateEmailSend, 120) = #dateEmailSend
AND B.[Company] = #ola_company
AND B.[Object Identified As] = #obj_identified_as
AND B.[Product Involved] = #product_involved
AND B.Region = #region
AND B.[Customer] = #ola_customer
AND A.CounterVal != #counter;
DECLARE #aCounter int;
BEGIN
SELECT
#dateEmailSend = CONVERT(varchar(10), A.dateEmailSend, 120),
#counter = B.Counter,
#region = B.Region,
#ola_company = B.[Company],
#obj_identified_as = B.[Object Identified As],
#product_involved = B.[Product Involved],
#ola_customer = B.[Customer],
#product_ut = CONVERT(varchar(10), B.[produt u/t], 120)
FROM [cmpEmailAlerted] A,
[cmpTblProductComplaints] B
WHERE A.CounterVal = #counterVal
AND a.CounterVal = B.Counter;
OPEN otherCounterVals;
FETCH NEXT FROM otherCounterVals INTO #aCounter;
PRINT ##fetch_status;
PRINT 'start of the loop';
WHILE ##FETCH_STATUS = 0
BEGIN
PRINT 'in the loop';
SELECT
#RESULTDESC = concat(#resultdesc, #aCounter, ',');
FETCH NEXT FROM otherCounterVals INTO #aCounter;
END;
PRINT 'out the loop';
CLOSE otherCounterVals;
DEALLOCATE otherCounterVals;
--remove the last comma from the resultdesc
IF (SUBSTRING(#RESULTDESC, LEN(#RESULTDESC), 1) = ',')
BEGIN
SELECT
#RESULTDESC = SUBSTRING(#RESULTDESC, 0, (LEN(#RESULTDESC)));
END
SELECT
#RESULT = 'SUCCESS';
END;
Here is the output i am getting:
-1 start of the loop out the loop
(1 row affected)
(1 row affected)

If you want to just build up a comma delimited string of results, there are much, much easier ways of doing it. A straight-forward, naïve way that doesn't rely on any magic STUFF functionality is simply this pattern:
DECLARE #result varchar(max);
SET #result = '';
SELECT #result = #result + ',' + ColumnValueToUse FROM MyTable WHERE <somecondition>
PRINT #result;
You can then peel off the first comma. The result will be every value of "ColumValueToUse" in the "MyTable" table that meets whatever the condition is.... all concatenated together and comma separated. There are fancier ways to do this with STUFF() and FOR XML that handle the comma thing better, and let you do even more fancy stuff.
It's ridiculous to use an expensive cursor (that isn't even declared FAST_FORWARD) for this. Please don't.
As for why your cursor solution isn't working, first ensure your SELECT statement returns the rows you want. Given you don't actually set any of the variables used by the SELECT in the cursor definition, it probably doesn't. But mostly just remove the cursor, please.
Trying to extrapolate from your own example, something like this:
DECLARE #result varchar(max);
SET #result = '';
SELECT #result = #result + ',' + A.CounterVal
FROM [cmpEmailAlerted] A,
[cmpTblProductComplaints] B
WHERE A.CounterVal = B.Counter
AND CONVERT(varchar(10), A.dateEmailSend, 120) = #dateEmailSend
AND B.[Company] = #ola_company
AND B.[Object Identified As] = #obj_identified_as
AND B.[Product Involved] = #product_involved
AND B.Region = #region
AND B.[Customer] = #ola_customer
AND A.CounterVal != #counter;
SELECT #result;

The # values are not yet defined for the first cursor listed.

You need an inner cursor per row of the SELECT statement as well as an outer cursor for the outer SELECT statement.

Related

How to iterate over a string of varying length, replacing different abbreviations with their full text. All abbreviations separated by a semicolon

My problem is this; I have a field in a table that contains values like this:
NP
NP;MC;PE
MC;AB;AT;MI;TC;WM
OS
OG
I want to convert these abbreviations to their full name. i.e. NP becomes Nuclear Power, OG becomes Oil and Gas, MI becomes Military etc.
My desired output would be:
Nuclear Power
Nuclear Power;Military;Pesticides
and so on.
I'm creating this as a function. I got it working for just the one abbreviation and then the same for two. However my issue is that I may have 5 abbreviations or 7. I know my current approach is dreadful but cannot figure out how to loop it in the right way.
Please note: I've shortened the list of abbreviations for StackOverflow but there's 25 in total.
Please further note: I did the function bottom up (I don't know why) and got the two value and single value working. I've removed anything I did for values over 3 as nothing I did worked.
ALTER FUNCTION [dbo].[get_str_full]
(
-- Add the parameters for the function here
#str_input VARCHAR(250)
)
RETURNS VARCHAR(250)
AS
BEGIN
-- Declare the return variable here
DECLARE #Result VARCHAR(250)
DECLARE #TEMPSTRING VARCHAR(250)
DECLARE #TEMPSTRING_RIGHT AS VARCHAR(250)
-- DECLARE #PI_COUNT BIGINT
DECLARE #COUNTER INT
DECLARE #TOTAL_VALS BIGINT
DECLARE #STRING_ST VARCHAR(250)
DECLARE #POS_STR BIGINT
DECLARE #REMAINING_STR VARCHAR(250)
-- Used for easy loop skips
DECLARE #LEFTSKIP AS BIGINT
SET #LEFTSKIP = 1
SET #Result = #str_input
SET #STRING_ST = #Result
SET #COUNTER = (LEN(#Result) - LEN(REPLACE(#Result,';',''))) + 1
SET #TOTAL_VALS = (LEN(#Result) - LEN(REPLACE(#Result,';',''))) + 1
-- If the string has a semicolon then there's more than one PI value
IF CHARINDEX(';', #Result) > 0
BEGIN
WHILE #COUNTER > 0
BEGIN
IF #TOTAL_VALS >= 3 -- If counter is more than 2 then there's three or more
BEGIN
DECLARE #TEMP_VAL BIGINT
SET #TEMP_VAL = 5
END
ELSE IF #TOTAL_VALS = 2-- Theres 2
BEGIN
-- Do left two chars first
IF #LEFTSKIP = 1
BEGIN
SET #TEMPSTRING = LEFT(#Result, 2)
SELECT #TEMPSTRING = CASE #TEMPSTRING
WHEN 'MC' THEN 'Military Contracting'
WHEN 'NP' THEN 'Nuclear'
WHEN 'OG' THEN 'Oil & Gas'
WHEN 'OS' THEN 'Oil Sands'
WHEN 'PM' THEN 'Palm Oil'
WHEN 'PE' THEN 'Pesticides'
ELSE #TEMPSTRING
END
SET #LEFTSKIP = 2
END
ELSE IF #LEFTSKIP = 2
BEGIN
SET #TEMPSTRING_RIGHT = RIGHT(#Result, 2)
SELECT #TEMPSTRING_RIGHT = CASE #TEMPSTRING_RIGHT
WHEN 'MC' THEN 'Military Contracting'
WHEN 'NP' THEN 'Nuclear'
WHEN 'OG' THEN 'Oil & Gas'
WHEN 'OS' THEN 'Oil Sands'
WHEN 'PM' THEN 'Palm Oil'
WHEN 'PE' THEN 'Pesticides'
ELSE #TEMPSTRING_RIGHT
END
END
END
SET #COUNTER = #COUNTER - 1
END
SET #Result = CONCAT(#TEMPSTRING,';', #TEMPSTRING_RIGHT)
END
ELSE
BEGIN
SET #Result = REPLACE(#Result, 'MC', 'Military Contracting')
SET #Result = REPLACE(#RESULT, 'NP', 'Nuclear Power')
SET #Result = REPLACE(#Result, 'OG', 'Oil & Gas')
SET #Result = REPLACE(#Result, 'OS', 'Oil Sands')
SET #Result = REPLACE(#Result, 'PM', 'Palm Oil')
SET #Result = REPLACE(#Result, 'PE', 'Pesticides')
END
-- Return the result of the function
RETURN #Result
END
First for some easily consumable sample data:
DECLARE #tranlation TABLE(tCode VARCHAR(10), tString VARCHAR(40));
DECLARE #t TABLE(String VARCHAR(1000));
INSERT #t VALUES('PE;N'),('NP'),('NP;MC;PE;XX')
INSERT #tranlation VALUES ('N','Nukes'),('NP','Nuclear Power'),('MC','Military'),
('PE','Pesticides');
Note my updated sample data which includes "XX", which has no match , and an "N" for "Nukes" which would wreck any solution which leverages REPLACE. If you are on SQL 2016+ you can use STRING_SPLIT and STRING_AGG.
SELECT
OldString = t.String,
NewString = STRING_AGG(ISNULL(tx.tString,items.[value]),';')
FROM #t AS t
OUTER APPLY STRING_SPLIT(t.String,';') AS items
LEFT JOIN #tranlation AS tx
ON items.[value] = tx.tCode
GROUP BY t.String ;
Returns:
OldString NewString
----------------- -------------------------------------------
NP Nuclear Power
NP;MC;PE;XX Nuclear Power;Military;Pesticides;XX
PE;N Pesticides;Nukes
You should really fix your table design so that you do not store multiple pieces of info in one column.
If you would like it as a function, I would strongly recommend an inline Table-Valued function rather than a scalar function.
If you have SQL Server version 2017+ you can use STRING_SPLIT and STRING_AGG for this.
CREATE OR ALTER FUNCTION GetFullStr
( #str varchar(250) )
RETURNS TABLE
AS RETURN
(
SELECT STRING_AGG(ISNULL(v.FullStr, s.value), ';') result
FROM STRING_SPLIT(#str, ';') s
LEFT JOIN (VALUES
('MC', 'Military Contracting'),
('NP', 'Nuclear'),
('OG', 'Oil & Gas'),
('OS', 'Oil Sands'),
('PM', 'Palm Oil'),
('PE', 'Pesticides')
) v(Abbr, FullStr) ON v.Abbr = s.value
);
GO
You can, and should, replace the VALUES with a real table.
On 2016 you would need FOR XML PATH instead of STRING_AGG:
CREATE OR ALTER FUNCTION GetFullStr
( #str varchar(250) )
RETURNS TABLE
AS RETURN
(
SELECT STUFF(
(SELECT ';' + ISNULL(v.FullStr, s.value)
FROM STRING_SPLIT(#str, ';') s
LEFT JOIN (VALUES
('MC', 'Military Contracting'),
('NP', 'Nuclear'),
('OG', 'Oil & Gas'),
('OS', 'Oil Sands'),
('PM', 'Palm Oil'),
('PE', 'Pesticides')
) v(Abbr, FullStr) ON v.Abbr = s.value
FOR XML PATH(''), TYPE
).value('text()[1]','varchar(2500)'),
, 1, 1, '')
);
GO
You use it like this:
SELECT s.result AS FullStr
FROM table
OUTER APPLY GetFullStr(value) AS s;
-- alternatively
SELECT (SELECT * FROM GetFullStr(value)) AS FullStr
FROM table;
You could assign your abbreviation mappings to a TABLE variable and then use that for your REPLACE. You could build this into a function, then pass your string values in.
The test below returns Military:Nuclear Power:XX.
declare #mapping table (abbrev varchar(50), fullname varchar(100))
insert into #mapping(abbrev, fullname)
values ('NP','Nuclear Power'),
('MC','Military')
declare #testString varchar(100), #newString varchar(100)
set #teststring = 'MC:NP:XX'
set #newString = #testString
SELECT #newString = REPLACE(#newString, abbrev, fullname) FROM #mapping
select #newString

SQL Server if statement does not execute as expected

I am trying to use the following stored procedure but there are some instances WHERE only the incremental happens AND the code does not run. What I need is that, when the program enters the IF statement, either it should run both the statements or None.
Stored procedure goes like this:
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
ALTER PROCEDURE [dbo].[spflpunch]
AS
BEGIN
DECLARE #id NUMERIC(18,0)
DECLARE #studname NVARCHAR(50)
DECLARE #punchtime DATETIME
DECLARE #samedaycount NUMERIC(2)
SELECT #id = (MAX(lastid)) FROM [smartswype].[dbo].[read]
PRINT #id
SELECT #studname = studname
FROM [SSWYPE_WEBDB].[dbo].[attdview]
WHERE id =#id
PRINT #studname
SELECT #punchtime = punchtime
FROM [SSWYPE_WEBDB].[dbo].[attdview]
WHERE id = #id
PRINT #punchtime
--SELECT #punchvarchar = CONVERT(VARCHAR(10),#punchtime, 103) + ' ' + CONVERT(VARCHAR(5), #punchtime, 14)
IF #id = (SELECT MAX(id) FROM [SSWYPE_WEBDB].[dbo].[attdview])
BEGIN
SELECT #samedaycount = COUNT(*)
FROM [SSWYPE_WEBDB].[dbo].[attdview]
WHERE (studname = #studname
AND CONVERT(DATE, punchtime) = CONVERT(DATE, #punchtime)) -- If firstpunch = 1 then it is the first punch
PRINT #samedaycount
IF #samedaycount =1
BEGIN
INSERT INTO [smartswype].[dbo].[firstlastpunch] ([studname], [DATE], [punch1], [punch2])
VALUES(#studname, CONVERT(DATE, #punchtime), #punchtime, NULL);
UPDATE [smartswype].[dbo].[read]
SET lastid = #id + 1;
END
ELSE IF (#samedaycount > 1)
BEGIN
UPDATE [smartswype].[dbo].[firstlastpunch]
SET punch2 = #punchtime
WHERE (studname = #studname AND DATE = CONVERT(DATE, #punchtime));
UPDATE [smartswype].[dbo].[read]
SET lastid = #id + 1;
END
END
END
If you want to ensure that both or none of the statements run, you should wrap the contents of the if statement in a transaction.
By wrapping it in a transaction, you can ensure that if one statement fails, that the other statement will not run.
Here is a link to the docs on transactions in SQL Server
https://learn.microsoft.com/en-us/sql/t-sql/language-elements/commit-transaction-transact-sql

Using Wildcards in SQL to delete part of a string [duplicate]

SELECT REPLACE('<strong>100</strong><b>.00 GB', '%^(^-?\d*\.{0,1}\d+$)%', '');
I want to replace any markup between two parts of the number with above regex, but it does not seem to work. I'm not sure if it is regex syntax that's wrong because I tried simpler one such as '%[^0-9]%' just to test but it didn't work either. Does anyone know how can I achieve this?
You can use PATINDEX
to find the first index of the pattern (string's) occurrence. Then use STUFF to stuff another string into the pattern(string) matched.
Loop through each row. Replace each illegal characters with what you want. In your case replace non numeric with blank. The inner loop is if you have more than one illegal character in a current cell that of the loop.
DECLARE #counter int
SET #counter = 0
WHILE(#counter < (SELECT MAX(ID_COLUMN) FROM Table))
BEGIN
WHILE 1 = 1
BEGIN
DECLARE #RetVal varchar(50)
SET #RetVal = (SELECT Column = STUFF(Column, PATINDEX('%[^0-9.]%', Column),1, '')
FROM Table
WHERE ID_COLUMN = #counter)
IF(#RetVal IS NOT NULL)
UPDATE Table SET
Column = #RetVal
WHERE ID_COLUMN = #counter
ELSE
break
END
SET #counter = #counter + 1
END
Caution: This is slow though! Having a varchar column may impact. So using LTRIM RTRIM may help a bit. Regardless, it is slow.
Credit goes to this StackOverFlow answer.
EDIT
Credit also goes to #srutzky
Edit (by #Tmdean)
Instead of doing one row at a time, this answer can be adapted to a more set-based solution. It still iterates the max of the number of non-numeric characters in a single row, so it's not ideal, but I think it should be acceptable in most situations.
WHILE 1 = 1 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, '')
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 BREAK;
END;
You can also improve efficiency quite a lot if you maintain a bit column in the table that indicates whether the field has been scrubbed yet. (NULL represents "Unknown" in my example and should be the column default.)
DECLARE #done bit = 0;
WHILE #done = 0 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table
WHERE COALESCE(Scrubbed_Column, 0) = 0)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, ''),
Scrubbed_Column = 0
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 SET #done = 1;
-- if Scrubbed_Column is still NULL, then the PATINDEX
-- must have given 0
UPDATE table
SET Scrubbed_Column = CASE
WHEN Scrubbed_Column IS NULL THEN 1
ELSE NULLIF(Scrubbed_Column, 0)
END;
END;
If you don't want to change your schema, this is easy to adapt to store intermediate results in a table valued variable which gets applied to the actual table at the end.
Instead of stripping out the found character by its sole position, using Replace(Column, BadFoundCharacter, '') could be substantially faster. Additionally, instead of just replacing the one bad character found next in each column, this replaces all those found.
WHILE 1 = 1 BEGIN
UPDATE dbo.YourTable
SET Column = Replace(Column, Substring(Column, PatIndex('%[^0-9.-]%', Column), 1), '')
WHERE Column LIKE '%[^0-9.-]%'
If ##RowCount = 0 BREAK;
END;
I am convinced this will work better than the accepted answer, if only because it does fewer operations. There are other ways that might also be faster, but I don't have time to explore those right now.
In a general sense, SQL Server does not support regular expressions and you cannot use them in the native T-SQL code.
You could write a CLR function to do that. See here, for example.
For those looking for a performant and easy solution and are willing to enable CLR:
CREATE database TestSQLFunctions
go
use TestSQLFunctions
go
ALTER database TestSQLFunctions set trustworthy on
EXEC sp_configure 'clr enabled', 1
RECONFIGURE WITH OVERRIDE
go
CREATE ASSEMBLY [SQLFunctions]
AUTHORIZATION [dbo]
FROM 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WITH PERMISSION_SET = SAFE
go
CREATE FUNCTION RegexReplace(
#input nvarchar(max),
#pattern nvarchar(max),
#replacement nvarchar(max)
) RETURNS nvarchar (max)
AS EXTERNAL NAME SQLFunctions.[SQLFunctions.Regex].Replace;
go
-- outputs This is a test
SELECT dbo.RegexReplace('This is a test 12345','[0-9]','')
Content of the DLL:
I stumbled across this post looking for something else but thought I'd mention a solution I use which is far more efficient - and really should be the default implementation of any function when used with a set based query - which is to use a cross applied table function. Seems the topic is still active so hopefully this is useful to someone.
Example runtime on a few of the answers so far based on running recursive set based queries or scalar function, based on 1m rows test set removing the chars from a random newid, ranges from 34s to 2m05s for the WHILE loop examples and from 1m3s to {forever} for the function examples.
Using a table function with cross apply achieves the same goal in 10s. You may need to adjust it to suit your needs such as the max length it handles.
Function:
CREATE FUNCTION [dbo].[RemoveChars](#InputUnit VARCHAR(40))
RETURNS TABLE
AS
RETURN
(
WITH Numbers_prep(Number) AS
(
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
)
,Numbers(Number) AS
(
SELECT TOP (ISNULL(LEN(#InputUnit),0))
row_number() OVER (ORDER BY (SELECT NULL))
FROM Numbers_prep a
CROSS JOIN Numbers_prep b
)
SELECT
OutputUnit
FROM
(
SELECT
substring(#InputUnit,Number,1)
FROM Numbers
WHERE substring(#InputUnit,Number,1) like '%[0-9]%'
ORDER BY Number
FOR XML PATH('')
) Sub(OutputUnit)
)
Usage:
UPDATE t
SET column = o.OutputUnit
FROM ##t t
CROSS APPLY [dbo].[RemoveChars](t.column) o
Here is a function I wrote to accomplish this based off of the previous answers.
CREATE FUNCTION dbo.RepetitiveReplace
(
#P_String VARCHAR(MAX),
#P_Pattern VARCHAR(MAX),
#P_ReplaceString VARCHAR(MAX),
#P_ReplaceLength INT = 1
)
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #Index INT;
-- Get starting point of pattern
SET #Index = PATINDEX(#P_Pattern, #P_String);
while #Index > 0
begin
--replace matching charactger at index
SET #P_String = STUFF(#P_String, PATINDEX(#P_Pattern, #P_String), #P_ReplaceLength, #P_ReplaceString);
SET #Index = PATINDEX(#P_Pattern, #P_String);
end
RETURN #P_String;
END;
[Gist][1]
[1]: https://gist.github.com/jkdba/ca13fe8f2a9855c4bdbfd0a5d3dfcda2
Edit:
Originally I had a recursive function here which does not play well with sql server as it has a 32 nesting level limit which would result in an error like the below any time you attempt to make 32+ replacements with the function. Instead of trying to make a server level change to allow more nesting (which could be dangerous like allow never ending loops) switching to a while loop makes a lot more sense.
Maximum stored procedure, function, trigger, or view nesting level exceeded (limit 32).
Wrapping the solution inside a SQL function could be useful if you want to reuse it.
I'm even doing it at the cell level, that's why I'm putting this as a different answer:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(300))
RETURNS VARCHAR(300)
BEGIN
DECLARE #str VARCHAR(300) = #string;
DECLARE #Pattern VARCHAR (20) = '%[^a-zA-Z0-9]%';
DECLARE #Len INT;
SELECT #Len = LEN(#String);
WHILE #Len > 0
BEGIN
SET #Len = #Len - 1;
IF (PATINDEX(#Pattern,#str) > 0)
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
A more speedy approach for large strings would look something like this:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(MAX))
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #str VARCHAR(MAX) = #string;
DECLARE #Pattern VARCHAR (MAX) = '%[^a-zA-Z0-9]%';
WHILE PATINDEX(#Pattern,#str) > 0
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
RETURN #str
END
I've created this function to clean up a string that contained non numeric characters in a time field. The time contained question marks when they did not added the minutes, something like this 20:??. Function loops through each character and replaces the ? with a 0 :
CREATE FUNCTION [dbo].[CleanTime]
(
-- Add the parameters for the function here
#intime nvarchar(10)
)
RETURNS nvarchar(5)
AS
BEGIN
-- Declare the return variable here
DECLARE #ResultVar nvarchar(5)
DECLARE #char char(1)
-- Add the T-SQL statements to compute the return value here
DECLARE #i int = 1
WHILE #i <= LEN(#intime)
BEGIN
SELECT #char = CASE WHEN substring(#intime,#i,1) like '%[0-9:]%' THEN substring(#intime,#i,1) ELSE '0' END
SELECT #ResultVar = concat(#ResultVar,#char)
set #i = #i + 1
END;
-- Return the result of the function
RETURN #ResultVar
END
I think this solution is faster and simple. I use always CTE/recursive because WHILE is so slow on SQL Server.
I use it in projects I work with and large databases.
/*
Function: dbo.kSql_ReplaceRegExp
Create Date: 20.02.2021
Author: Karcan Ozbal
Description: The given string value will be replaced according to the given regexp/pattern.
Parameter(s): #Value : Value/Text to REPLACE.
#RegExp : The regexp/pattern to be used for REPLACE operation.
Usage: select dbo.kSql_ReplaceRegExp('2T3EST5','%[0-9]%')
Output: 'TEST'
*/
ALTER FUNCTION [dbo].[kSql_ReplaceRegExp](
#Value nvarchar(max),
#RegExp nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #Result nvarchar(max)
;WITH CTE AS (
SELECT NUM = 1, VALUE = #Value, IDX = PATINDEX(#RegExp, #Value)
UNION ALL
SELECT NUM + 1, VALUE = REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''), IDX = PATINDEX(#RegExp, REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''))
FROM CTE
WHERE IDX > 0
)
SELECT TOP(1) #Result = VALUE
FROM CTE
ORDER BY NUM DESC
OPTION (maxrecursion 0)
RETURN #Result
END
If you are doing this just for a parameter coming into a Stored Procedure, you can use the following:
declare #badIndex int
set #badIndex = PatIndex('%[^0-9]%', #Param)
while #badIndex > 0
set #Param = Replace(#Param, Substring(#Param, #badIndex, 1), '')
set #badIndex = PatIndex('%[^0-9]%', #Param)
I thought this was clearer:
ALTER FUNCTION [dbo].[func_ReplaceChars](
#Value nvarchar(max),
#Chars nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #cLen int = len(#Chars);
DECLARE #curChar int = 0;
WHILE #curChar<#cLen
BEGIN
set #Value = replace(#Value,substring(#Chars,#curChar,1),'');
set #curChar = #curChar + 1;
END;
RETURN #Value
END
I'm using this code similar to several codes above:
DROP FUNCTION [dbo].[fnCleanString]
GO
CREATE FUNCTION [dbo].[fnCleanString] (#input VARCHAR(max), #Pattern
VARCHAR (20))
RETURNS VARCHAR(max)
BEGIN
DECLARE #str VARCHAR(max) = #input;
DECLARE #Len INT;
DECLARE #INDEX INT;
SELECT #Len = LEN(#input);
WHILE #Len > 0
BEGIN
SET #INDEX = PATINDEX(#Pattern,#str);
IF (#INDEX > 0)
BEGIN
SET #str=REPLACE(#str,SUBSTRING(#str,#INDEX, 1), '');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
You can use it like this:
SELECT CleanName = dbo.[fnCleanString](Name, '%[0-9]%') from YourTable
I think a simpler and faster approach is iterate by each character of the alphabet:
DECLARE #i int
SET #i = 0
WHILE(#i < 256)
BEGIN
IF char(#i) NOT IN ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.')
UPDATE Table SET Column = replace(Column, char(#i), '')
SET #i = #i + 1
END

Regex pattern inside SQL Replace function?

SELECT REPLACE('<strong>100</strong><b>.00 GB', '%^(^-?\d*\.{0,1}\d+$)%', '');
I want to replace any markup between two parts of the number with above regex, but it does not seem to work. I'm not sure if it is regex syntax that's wrong because I tried simpler one such as '%[^0-9]%' just to test but it didn't work either. Does anyone know how can I achieve this?
You can use PATINDEX
to find the first index of the pattern (string's) occurrence. Then use STUFF to stuff another string into the pattern(string) matched.
Loop through each row. Replace each illegal characters with what you want. In your case replace non numeric with blank. The inner loop is if you have more than one illegal character in a current cell that of the loop.
DECLARE #counter int
SET #counter = 0
WHILE(#counter < (SELECT MAX(ID_COLUMN) FROM Table))
BEGIN
WHILE 1 = 1
BEGIN
DECLARE #RetVal varchar(50)
SET #RetVal = (SELECT Column = STUFF(Column, PATINDEX('%[^0-9.]%', Column),1, '')
FROM Table
WHERE ID_COLUMN = #counter)
IF(#RetVal IS NOT NULL)
UPDATE Table SET
Column = #RetVal
WHERE ID_COLUMN = #counter
ELSE
break
END
SET #counter = #counter + 1
END
Caution: This is slow though! Having a varchar column may impact. So using LTRIM RTRIM may help a bit. Regardless, it is slow.
Credit goes to this StackOverFlow answer.
EDIT
Credit also goes to #srutzky
Edit (by #Tmdean)
Instead of doing one row at a time, this answer can be adapted to a more set-based solution. It still iterates the max of the number of non-numeric characters in a single row, so it's not ideal, but I think it should be acceptable in most situations.
WHILE 1 = 1 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, '')
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 BREAK;
END;
You can also improve efficiency quite a lot if you maintain a bit column in the table that indicates whether the field has been scrubbed yet. (NULL represents "Unknown" in my example and should be the column default.)
DECLARE #done bit = 0;
WHILE #done = 0 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table
WHERE COALESCE(Scrubbed_Column, 0) = 0)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, ''),
Scrubbed_Column = 0
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 SET #done = 1;
-- if Scrubbed_Column is still NULL, then the PATINDEX
-- must have given 0
UPDATE table
SET Scrubbed_Column = CASE
WHEN Scrubbed_Column IS NULL THEN 1
ELSE NULLIF(Scrubbed_Column, 0)
END;
END;
If you don't want to change your schema, this is easy to adapt to store intermediate results in a table valued variable which gets applied to the actual table at the end.
Instead of stripping out the found character by its sole position, using Replace(Column, BadFoundCharacter, '') could be substantially faster. Additionally, instead of just replacing the one bad character found next in each column, this replaces all those found.
WHILE 1 = 1 BEGIN
UPDATE dbo.YourTable
SET Column = Replace(Column, Substring(Column, PatIndex('%[^0-9.-]%', Column), 1), '')
WHERE Column LIKE '%[^0-9.-]%'
If ##RowCount = 0 BREAK;
END;
I am convinced this will work better than the accepted answer, if only because it does fewer operations. There are other ways that might also be faster, but I don't have time to explore those right now.
In a general sense, SQL Server does not support regular expressions and you cannot use them in the native T-SQL code.
You could write a CLR function to do that. See here, for example.
For those looking for a performant and easy solution and are willing to enable CLR:
CREATE database TestSQLFunctions
go
use TestSQLFunctions
go
ALTER database TestSQLFunctions set trustworthy on
EXEC sp_configure 'clr enabled', 1
RECONFIGURE WITH OVERRIDE
go
CREATE ASSEMBLY [SQLFunctions]
AUTHORIZATION [dbo]
FROM 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
WITH PERMISSION_SET = SAFE
go
CREATE FUNCTION RegexReplace(
#input nvarchar(max),
#pattern nvarchar(max),
#replacement nvarchar(max)
) RETURNS nvarchar (max)
AS EXTERNAL NAME SQLFunctions.[SQLFunctions.Regex].Replace;
go
-- outputs This is a test
SELECT dbo.RegexReplace('This is a test 12345','[0-9]','')
Content of the DLL:
I stumbled across this post looking for something else but thought I'd mention a solution I use which is far more efficient - and really should be the default implementation of any function when used with a set based query - which is to use a cross applied table function. Seems the topic is still active so hopefully this is useful to someone.
Example runtime on a few of the answers so far based on running recursive set based queries or scalar function, based on 1m rows test set removing the chars from a random newid, ranges from 34s to 2m05s for the WHILE loop examples and from 1m3s to {forever} for the function examples.
Using a table function with cross apply achieves the same goal in 10s. You may need to adjust it to suit your needs such as the max length it handles.
Function:
CREATE FUNCTION [dbo].[RemoveChars](#InputUnit VARCHAR(40))
RETURNS TABLE
AS
RETURN
(
WITH Numbers_prep(Number) AS
(
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
)
,Numbers(Number) AS
(
SELECT TOP (ISNULL(LEN(#InputUnit),0))
row_number() OVER (ORDER BY (SELECT NULL))
FROM Numbers_prep a
CROSS JOIN Numbers_prep b
)
SELECT
OutputUnit
FROM
(
SELECT
substring(#InputUnit,Number,1)
FROM Numbers
WHERE substring(#InputUnit,Number,1) like '%[0-9]%'
ORDER BY Number
FOR XML PATH('')
) Sub(OutputUnit)
)
Usage:
UPDATE t
SET column = o.OutputUnit
FROM ##t t
CROSS APPLY [dbo].[RemoveChars](t.column) o
Here is a function I wrote to accomplish this based off of the previous answers.
CREATE FUNCTION dbo.RepetitiveReplace
(
#P_String VARCHAR(MAX),
#P_Pattern VARCHAR(MAX),
#P_ReplaceString VARCHAR(MAX),
#P_ReplaceLength INT = 1
)
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #Index INT;
-- Get starting point of pattern
SET #Index = PATINDEX(#P_Pattern, #P_String);
while #Index > 0
begin
--replace matching charactger at index
SET #P_String = STUFF(#P_String, PATINDEX(#P_Pattern, #P_String), #P_ReplaceLength, #P_ReplaceString);
SET #Index = PATINDEX(#P_Pattern, #P_String);
end
RETURN #P_String;
END;
[Gist][1]
[1]: https://gist.github.com/jkdba/ca13fe8f2a9855c4bdbfd0a5d3dfcda2
Edit:
Originally I had a recursive function here which does not play well with sql server as it has a 32 nesting level limit which would result in an error like the below any time you attempt to make 32+ replacements with the function. Instead of trying to make a server level change to allow more nesting (which could be dangerous like allow never ending loops) switching to a while loop makes a lot more sense.
Maximum stored procedure, function, trigger, or view nesting level exceeded (limit 32).
Wrapping the solution inside a SQL function could be useful if you want to reuse it.
I'm even doing it at the cell level, that's why I'm putting this as a different answer:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(300))
RETURNS VARCHAR(300)
BEGIN
DECLARE #str VARCHAR(300) = #string;
DECLARE #Pattern VARCHAR (20) = '%[^a-zA-Z0-9]%';
DECLARE #Len INT;
SELECT #Len = LEN(#String);
WHILE #Len > 0
BEGIN
SET #Len = #Len - 1;
IF (PATINDEX(#Pattern,#str) > 0)
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
A more speedy approach for large strings would look something like this:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(MAX))
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #str VARCHAR(MAX) = #string;
DECLARE #Pattern VARCHAR (MAX) = '%[^a-zA-Z0-9]%';
WHILE PATINDEX(#Pattern,#str) > 0
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
RETURN #str
END
I've created this function to clean up a string that contained non numeric characters in a time field. The time contained question marks when they did not added the minutes, something like this 20:??. Function loops through each character and replaces the ? with a 0 :
CREATE FUNCTION [dbo].[CleanTime]
(
-- Add the parameters for the function here
#intime nvarchar(10)
)
RETURNS nvarchar(5)
AS
BEGIN
-- Declare the return variable here
DECLARE #ResultVar nvarchar(5)
DECLARE #char char(1)
-- Add the T-SQL statements to compute the return value here
DECLARE #i int = 1
WHILE #i <= LEN(#intime)
BEGIN
SELECT #char = CASE WHEN substring(#intime,#i,1) like '%[0-9:]%' THEN substring(#intime,#i,1) ELSE '0' END
SELECT #ResultVar = concat(#ResultVar,#char)
set #i = #i + 1
END;
-- Return the result of the function
RETURN #ResultVar
END
I think this solution is faster and simple. I use always CTE/recursive because WHILE is so slow on SQL Server.
I use it in projects I work with and large databases.
/*
Function: dbo.kSql_ReplaceRegExp
Create Date: 20.02.2021
Author: Karcan Ozbal
Description: The given string value will be replaced according to the given regexp/pattern.
Parameter(s): #Value : Value/Text to REPLACE.
#RegExp : The regexp/pattern to be used for REPLACE operation.
Usage: select dbo.kSql_ReplaceRegExp('2T3EST5','%[0-9]%')
Output: 'TEST'
*/
ALTER FUNCTION [dbo].[kSql_ReplaceRegExp](
#Value nvarchar(max),
#RegExp nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #Result nvarchar(max)
;WITH CTE AS (
SELECT NUM = 1, VALUE = #Value, IDX = PATINDEX(#RegExp, #Value)
UNION ALL
SELECT NUM + 1, VALUE = REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''), IDX = PATINDEX(#RegExp, REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''))
FROM CTE
WHERE IDX > 0
)
SELECT TOP(1) #Result = VALUE
FROM CTE
ORDER BY NUM DESC
OPTION (maxrecursion 0)
RETURN #Result
END
If you are doing this just for a parameter coming into a Stored Procedure, you can use the following:
declare #badIndex int
set #badIndex = PatIndex('%[^0-9]%', #Param)
while #badIndex > 0
set #Param = Replace(#Param, Substring(#Param, #badIndex, 1), '')
set #badIndex = PatIndex('%[^0-9]%', #Param)
I thought this was clearer:
ALTER FUNCTION [dbo].[func_ReplaceChars](
#Value nvarchar(max),
#Chars nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #cLen int = len(#Chars);
DECLARE #curChar int = 0;
WHILE #curChar<#cLen
BEGIN
set #Value = replace(#Value,substring(#Chars,#curChar,1),'');
set #curChar = #curChar + 1;
END;
RETURN #Value
END
I'm using this code similar to several codes above:
DROP FUNCTION [dbo].[fnCleanString]
GO
CREATE FUNCTION [dbo].[fnCleanString] (#input VARCHAR(max), #Pattern
VARCHAR (20))
RETURNS VARCHAR(max)
BEGIN
DECLARE #str VARCHAR(max) = #input;
DECLARE #Len INT;
DECLARE #INDEX INT;
SELECT #Len = LEN(#input);
WHILE #Len > 0
BEGIN
SET #INDEX = PATINDEX(#Pattern,#str);
IF (#INDEX > 0)
BEGIN
SET #str=REPLACE(#str,SUBSTRING(#str,#INDEX, 1), '');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
You can use it like this:
SELECT CleanName = dbo.[fnCleanString](Name, '%[0-9]%') from YourTable
I think a simpler and faster approach is iterate by each character of the alphabet:
DECLARE #i int
SET #i = 0
WHILE(#i < 256)
BEGIN
IF char(#i) NOT IN ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.')
UPDATE Table SET Column = replace(Column, char(#i), '')
SET #i = #i + 1
END

How do you count the number of occurrences of a certain substring in a SQL varchar?

I have a column that has values formatted like a,b,c,d. Is there a way to count the number of commas in that value in T-SQL?
The first way that comes to mind is to do it indirectly by replacing the comma with an empty string and comparing the lengths
Declare #string varchar(1000)
Set #string = 'a,b,c,d'
select len(#string) - len(replace(#string, ',', ''))
Quick extension of cmsjr's answer that works for strings with more than one character.
CREATE FUNCTION dbo.CountOccurrencesOfString
(
#searchString nvarchar(max),
#searchTerm nvarchar(max)
)
RETURNS INT
AS
BEGIN
return (LEN(#searchString)-LEN(REPLACE(#searchString,#searchTerm,'')))/LEN(#searchTerm)
END
Usage:
SELECT * FROM MyTable
where dbo.CountOccurrencesOfString(MyColumn, 'MyString') = 1
You can compare the length of the string with one where the commas are removed:
len(value) - len(replace(value,',',''))
The answer by #csmjr has a problem in some instances.
His answer was to do this:
Declare #string varchar(1000)
Set #string = 'a,b,c,d'
select len(#string) - len(replace(#string, ',', ''))
This works in most scenarios, however, try running this:
DECLARE #string VARCHAR(1000)
SET #string = 'a,b,c,d ,'
SELECT LEN(#string) - LEN(REPLACE(#string, ',', ''))
For some reason, REPLACE gets rid of the final comma but ALSO the space just before it (not sure why). This results in a returned value of 5 when you'd expect 4. Here is another way to do this which will work even in this special scenario:
DECLARE #string VARCHAR(1000)
SET #string = 'a,b,c,d ,'
SELECT LEN(REPLACE(#string, ',', '**')) - LEN(#string)
Note that you don't need to use asterisks. Any two-character replacement will do. The idea is that you lengthen the string by one character for each instance of the character you're counting, then subtract the length of the original. It's basically the opposite method of the original answer which doesn't come with the strange trimming side-effect.
Building on #Andrew's solution, you'll get much better performance using a non-procedural table-valued-function and CROSS APPLY:
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
/* Usage:
SELECT t.[YourColumn], c.StringCount
FROM YourDatabase.dbo.YourTable t
CROSS APPLY dbo.CountOccurrencesOfString('your search string', t.[YourColumn]) c
*/
CREATE FUNCTION [dbo].[CountOccurrencesOfString]
(
#searchTerm nvarchar(max),
#searchString nvarchar(max)
)
RETURNS TABLE
AS
RETURN
SELECT (DATALENGTH(#searchString)-DATALENGTH(REPLACE(#searchString,#searchTerm,'')))/NULLIF(DATALENGTH(#searchTerm), 0) AS StringCount
Declare #string varchar(1000)
DECLARE #SearchString varchar(100)
Set #string = 'as as df df as as as'
SET #SearchString = 'as'
select ((len(#string) - len(replace(#string, #SearchString, ''))) -(len(#string) -
len(replace(#string, #SearchString, ''))) % 2) / len(#SearchString)
Accepted answer is correct ,
extending it to use 2 or more character in substring:
Declare #string varchar(1000)
Set #string = 'aa,bb,cc,dd'
Set #substring = 'aa'
select (len(#string) - len(replace(#string, #substring, '')))/len(#substring)
Darrel Lee I think has a pretty good answer. Replace CHARINDEX() with PATINDEX(), and you can do some weak regex searching along a string, too...
Like, say you use this for #pattern:
set #pattern='%[-.|!,'+char(9)+']%'
Why would you maybe want to do something crazy like this?
Say you're loading delimited text strings into a staging table, where the field holding the data is something like a varchar(8000) or nvarchar(max)...
Sometimes it's easier/faster to do ELT (Extract-Load-Transform) with data rather than ETL (Extract-Transform-Load), and one way to do this is to load the delimited records as-is into a staging table, especially if you may want an simpler way to see the exceptional records rather than deal with them as part of an SSIS package...but that's a holy war for a different thread.
If we know there is a limitation on LEN and space, why cant we replace the space first?
Then we know there is no space to confuse LEN.
len(replace(#string, ' ', '-')) - len(replace(replace(#string, ' ', '-'), ',', ''))
Use this code, it is working perfectly.
I have create a sql function that accept two parameters, the first param is the long string that we want to search into it,and it can accept string length up to 1500 character(of course you can extend it or even change it to text datatype).
And the second parameter is the substring that we want to calculate the number of its occurance(its length is up to 200 character, of course you can change it to what your need). and the output is an integer, represent the number of frequency.....enjoy it.
CREATE FUNCTION [dbo].[GetSubstringCount]
(
#InputString nvarchar(1500),
#SubString NVARCHAR(200)
)
RETURNS int
AS
BEGIN
declare #K int , #StrLen int , #Count int , #SubStrLen int
set #SubStrLen = (select len(#SubString))
set #Count = 0
Set #k = 1
set #StrLen =(select len(#InputString))
While #K <= #StrLen
Begin
if ((select substring(#InputString, #K, #SubStrLen)) = #SubString)
begin
if ((select CHARINDEX(#SubString ,#InputString)) > 0)
begin
set #Count = #Count +1
end
end
Set #K=#k+1
end
return #Count
end
In SQL 2017 or higher, you can use this:
declare #hits int = 0
set #hits = (select value from STRING_SPLIT('F609,4DFA,8499',','));
select count(#hits)
Improved version based on top answer and other answers:
Wrapping the string with delimiters ensures that LEN works properly. Making the replace character string one character longer than the match string removes the need for division.
CREATE FUNCTION dbo.MatchCount(#value nvarchar(max), #match nvarchar(max))
RETURNS int
BEGIN
RETURN LEN('[' + REPLACE(#value,#match,REPLICATE('*', LEN('[' + #match + ']') - 1)) + ']') - LEN('['+#value+']')
END
DECLARE #records varchar(400)
SELECT #records = 'a,b,c,d'
select LEN(#records) as 'Before removing Commas' , LEN(#records) - LEN(REPLACE(#records, ',', '')) 'After Removing Commans'
The following should do the trick for both single character and multiple character searches:
CREATE FUNCTION dbo.CountOccurrences
(
#SearchString VARCHAR(1000),
#SearchFor VARCHAR(1000)
)
RETURNS TABLE
AS
RETURN (
SELECT COUNT(*) AS Occurrences
FROM (
SELECT ROW_NUMBER() OVER (ORDER BY O.object_id) AS n
FROM sys.objects AS O
) AS N
JOIN (
VALUES (#SearchString)
) AS S (SearchString)
ON
SUBSTRING(S.SearchString, N.n, LEN(#SearchFor)) = #SearchFor
);
GO
---------------------------------------------------------------------------------------
-- Test the function for single and multiple character searches
---------------------------------------------------------------------------------------
DECLARE #SearchForComma VARCHAR(10) = ',',
#SearchForCharacters VARCHAR(10) = 'de';
DECLARE #TestTable TABLE
(
TestData VARCHAR(30) NOT NULL
);
INSERT INTO #TestTable
(
TestData
)
VALUES
('a,b,c,de,de ,d e'),
('abc,de,hijk,,'),
(',,a,b,cde,,');
SELECT TT.TestData,
CO.Occurrences AS CommaOccurrences,
CO2.Occurrences AS CharacterOccurrences
FROM #TestTable AS TT
OUTER APPLY dbo.CountOccurrences(TT.TestData, #SearchForComma) AS CO
OUTER APPLY dbo.CountOccurrences(TT.TestData, #SearchForCharacters) AS CO2;
The function can be simplified a bit using a table of numbers (dbo.Nums):
RETURN (
SELECT COUNT(*) AS Occurrences
FROM dbo.Nums AS N
JOIN (
VALUES (#SearchString)
) AS S (SearchString)
ON
SUBSTRING(S.SearchString, N.n, LEN(#SearchFor)) = #SearchFor
);
I finally write this function that should cover all the possible situations, adding a char prefix and suffix to the input. this char is evaluated to be different to any of the char conteined in the search parameter, so it can't affect the result.
CREATE FUNCTION [dbo].[CountOccurrency]
(
#Input nvarchar(max),
#Search nvarchar(max)
)
RETURNS int AS
BEGIN
declare #SearhLength as int = len('-' + #Search + '-') -2;
declare #conteinerIndex as int = 255;
declare #conteiner as char(1) = char(#conteinerIndex);
WHILE ((CHARINDEX(#conteiner, #Search)>0) and (#conteinerIndex>0))
BEGIN
set #conteinerIndex = #conteinerIndex-1;
set #conteiner = char(#conteinerIndex);
END;
set #Input = #conteiner + #Input + #conteiner
RETURN (len(#Input) - len(replace(#Input, #Search, ''))) / #SearhLength
END
usage
select dbo.CountOccurrency('a,b,c,d ,', ',')
Declare #MainStr nvarchar(200)
Declare #SubStr nvarchar(10)
Set #MainStr = 'nikhildfdfdfuzxsznikhilweszxnikhil'
Set #SubStr = 'nikhil'
Select (Len(#MainStr) - Len(REPLACE(#MainStr,#SubStr,'')))/Len(#SubStr)
this T-SQL code finds and prints all occurrences of pattern #p in sentence #s. you can do any processing on the sentence afterward.
declare #old_hit int = 0
declare #hit int = 0
declare #i int = 0
declare #s varchar(max)='alibcalirezaalivisualization'
declare #p varchar(max)='ali'
while #i<len(#s)
begin
set #hit=charindex(#p,#s,#i)
if #hit>#old_hit
begin
set #old_hit =#hit
set #i=#hit+1
print #hit
end
else
break
end
the result is:
1
6
13
20
I ended up using a CTE table for this,
CREATE TABLE #test (
[id] int,
[field] nvarchar(500)
)
INSERT INTO #test ([id], [field])
VALUES (1, 'this is a test string http://url, and https://google.com'),
(2, 'another string, hello world http://example.com'),
(3, 'a string with no url')
SELECT *
FROM #test
;WITH URL_count_cte ([id], [url_index], [field])
AS
(
SELECT [id], CHARINDEX('http', [field], 0)+1 AS [url_index], [field]
FROM #test AS [t]
WHERE CHARINDEX('http', [field], 0) != 0
UNION ALL
SELECT [id], CHARINDEX('http', [field], [url_index])+1 AS [url_index], [field]
FROM URL_count_cte
WHERE CHARINDEX('http', [field], [url_index]) > 0
)
-- total urls
SELECT COUNT(1)
FROM URL_count_cte
-- urls per row
SELECT [id], COUNT(1) AS [url_count]
FROM URL_count_cte
GROUP BY [id]
Using this function, you can get the number of repetitions of words in a text.
/****** Object: UserDefinedFunction [dbo].[fn_getCountKeywords] Script Date: 22/11/2021 17:52:00 ******/
DROP FUNCTION IF EXISTS [dbo].[fn_getCountKeywords]
GO
/****** Object: UserDefinedFunction [dbo].[fn_getCountKeywords] Script Date: 2211/2021 17:52:00 ******/
SET ANSI_NULLS OFF
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: m_Khezrian
-- Create date: 2021/11/22-17:52
-- Description: Return Count Keywords In Input Text
-- =============================================
Create OR Alter Function [dbo].[fn_getCountKeywords]
(#Text nvarchar(max)
,#Keywords nvarchar(max)
)
RETURNS #Result TABLE
(
[ID] int Not Null IDENTITY PRIMARY KEY
,[Keyword] nvarchar(max) Not Null
,[Cnt] int Not Null Default(0)
)
/*With ENCRYPTION*/ As
Begin
Declare #Key nvarchar(max);
Declare #Cnt int;
Declare #I int;
Set #I = 0 ;
--Set #Text = QUOTENAME(#Text);
Insert Into #Result
([Keyword])
Select Trim([value])
From String_Split(#Keywords,N',')
Group By [value]
Order By Len([value]) Desc;
Declare CntKey_Cursor Insensitive Cursor For
Select [Keyword]
From #Result
Order By [ID];
Open CntKey_Cursor;
Fetch Next From CntKey_Cursor Into #Key;
While (##Fetch_STATUS = 0) Begin
Set #Cnt = 0;
While (PatIndex(N'%'+#Key+'%',#Text) > 0) Begin
Set #Cnt += 1;
Set #I += 1 ;
Set #Text = Stuff(#Text,PatIndex(N'%'+#Key+'%',#Text),len(#Key),N'{'+Convert(nvarchar,#I)+'}');
--Set #Text = Replace(#Text,#Key,N'{'+Convert(nvarchar,#I)+'}');
End--While
Update #Result
Set [Cnt] = #Cnt
Where ([Keyword] = #Key);
Fetch Next From CntKey_Cursor Into #Key;
End--While
Close CntKey_Cursor;
Deallocate CntKey_Cursor;
Return
End
GO
--Test
Select *
From dbo.fn_getCountKeywords(
N'<U+0001F4E3> MARKET IMPACT Euro area Euro CPIarea annual inflation up to 3.0% MaCPIRKET forex'
,N'CPI ,core,MaRKET , Euro area'
)
Go
Reference https://learn.microsoft.com/en-us/sql/t-sql/functions/string-split-transact-sql?view=sql-server-ver15
Example:
SELECT s.*
,s.[Number1] - (SELECT COUNT(Value)
FROM string_split(s.[StringColumn],',')
WHERE RTRIM(VALUE) <> '')
FROM TableName AS s
Applies to: SQL Server 2016 (13.x) and later
You can use the following stored procedure to fetch , values.
IF EXISTS (SELECT * FROM sys.objects
WHERE object_id = OBJECT_ID(N'[dbo].[sp_parsedata]') AND type in (N'P', N'PC'))
DROP PROCEDURE [dbo].[sp_parsedata]
GO
create procedure sp_parsedata
(#cid integer,#st varchar(1000))
as
declare #coid integer
declare #c integer
declare #c1 integer
select #c1=len(#st) - len(replace(#st, ',', ''))
set #c=0
delete from table1 where complainid=#cid;
while (#c<=#c1)
begin
if (#c<#c1)
begin
select #coid=cast(replace(left(#st,CHARINDEX(',',#st,1)),',','') as integer)
select #st=SUBSTRING(#st,CHARINDEX(',',#st,1)+1,LEN(#st))
end
else
begin
select #coid=cast(#st as integer)
end
insert into table1(complainid,courtid) values(#cid,#coid)
set #c=#c+1
end
The Replace/Len test is cute, but probably very inefficient (especially in terms of memory).
A simple function with a loop will do the job.
CREATE FUNCTION [dbo].[fn_Occurences]
(
#pattern varchar(255),
#expression varchar(max)
)
RETURNS int
AS
BEGIN
DECLARE #Result int = 0;
DECLARE #index BigInt = 0
DECLARE #patLen int = len(#pattern)
SET #index = CHARINDEX(#pattern, #expression, #index)
While #index > 0
BEGIN
SET #Result = #Result + 1;
SET #index = CHARINDEX(#pattern, #expression, #index + #patLen)
END
RETURN #Result
END
Perhaps you should not store data that way. It is a bad practice to ever store a comma delimited list in a field. IT is very inefficient for querying. This should be a related table.

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