Related
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
To begin - I am new at SQL; be gentle.
I work with a school district and have recently been given "the keys" to access the database. I am interested in getting a list of students, and then generating a list of passwords for them. I have found some code which allows me to generate random passwords that I would like to incorporate into a SQL query which is gathering information from our Student database. (Thank you if this code is yours!)
My issue is that I have not been able to use the variables to create a different password for each record. I get the same randomly generated password for each student. (On a good note; at least the password changes each time I execute the query.)
I should mention that I have two accounts set up for the database access; one to simply read the information, and another with full editing rights. (I have only used this once and closed my eyes as I pushed the big red button to update some trivial information).
Results from the first run:
Name Password
JACEK mtwsz2ybu
CARL mtwsz2ybu
LARS mtwsz2ybu
Results from the second run:
Name Password
JACEK je4tm5ptw
CARL je4tm5ptw
LARS je4tm5ptw
This is the query I am running:
USE XXTableXX
DECLARE #position int, #string char(100), #length int,
#rand int, #newstring char(15), #newchar char(15);
SET #position = 1;
SET #string = 'abcdefghijkmnopqrstuvwxyz23456789';
SET #length = 9;
SET #newstring = ''
SET #newchar = ''
WHILE #position <= #length
BEGIN
SET #rand = FLOOR(RAND()*(33-1)+1);
SET #newchar = SUBSTRING(#string,#rand,1);
SET #newstring = STUFF(#newstring,len(#newstring)+1,1,#newchar)
SET #position = #position +1;
END;
SELECT DISTINCT s.firstname, #newstring AS [Password]
FROM XXStudentTableXX s
Put all the stuff above the query in a function and call the function in your query. A side note, it will execute for each row, which means a bit of a performance hit for large sets.
Here is a way to do this.
First, you will need to make the RAND() into it's own view since you can't call it from a function
CREATE VIEW [dbo].[NewRandom]
AS
SELECT RAND() AS [RandSeed]
GO
Now you must create your function that uses the view. This function accepts an integer so you can do variable password lengths. You can hardcode it in your query or take out the parameter and hard code it in the function.
CREATE FUNCTION [dbo].[ufn_GeneratePassword] ( #PasswordLength INT )
RETURNS VARCHAR(20)
AS
BEGIN
DECLARE #position int, #string char(100), #length int,
#rand int, #newstring char(15), #newchar char(15);
SET #position = 1;
SET #string = 'abcdefghijkmnopqrstuvwxyz23456789';
SET #length = #PasswordLength;
SET #newstring = ''
SET #newchar = ''
WHILE #position <= #length
BEGIN
SET #rand = FLOOR((SELECT RandSeed FROM dbo.[NewRandom])*(33-1)+1);
SET #newchar = SUBSTRING(#string,#rand,1);
SET #newstring = STUFF(#newstring,len(#newstring)+1,1,#newchar)
SET #position = #position +1;
END
RETURN #newstring
END
Now you can call the function on every row of your table
SELECT DISTINCT s.firstname, [dbo].[ufn_GeneratePassword](9) AS [Password]
FROM XXStudentTableXX s
I think what you will need to do for this is create a table, or table variable, then instead of selecting make it insert into the table. Also Add a loop that loops 1x for each student ID. This will then insert 1 row for each student and run the randomizer 1 time.
First covert your procedure into a scalar SQL function.
CREATE VIEW dbo.RandomNumberView
AS
SELECT RandomNumber = RAND();
GO
CREATE FUNCTION dbo.GeneratePassword()
RETURNS char(15)
AS
-- Generates and Returns a random password
BEGIN
DECLARE #position int, #string char(100), #length int,
#rand int, #newstring char(15), #newchar char(15);
SET #position = 1;
SET #string = 'abcdefghijkmnopqrstuvwxyz23456789';
SET #length = 9;
SET #newstring = ''
SET #newchar = ''
WHILE #position <= #length
BEGIN
SELECT #rand = FLOOR(RandomNumber *(33-1)+1) FROM RandomNumberView;
SET #newchar = SUBSTRING(#string,#rand,1);
SET #newstring = STUFF(#newstring,len(#newstring)+1,1,#newchar)
SET #position = #position +1;
END;
RETURN #newstring;
END;
GO
And then you can easily use it whatever way you want.
SELECT DISTINCT s.firstname, dbo.GeneratePassword() AS [Password]
FROM XXStudentTableXX s
Note that each time you run this query you would get different password against the same record. I would prefer to use this function to only generate and insert password in the table and not as a direct select statement.
To get the data from a random table, SQL Server provides the NEWID () function. As its name says every execution is generated a new GUID (Global Unique Identifier), these GUIDs are unique, so the value of the order will never be the same.
Therefore you do not need to create a random function that does it for you. So you may have your answer using a simple SQL.
CREATE TABLE STUDENT (
[ID] INT IDENTITY (1, 1) NOT NULL,
[NAME] NVARCHAR (100) NULL
);
INSERT INTO STUDENT (NAME) VALUES ('John');
INSERT INTO STUDENT (NAME) VALUES ('Carl');
INSERT INTO STUDENT (NAME) VALUES ('Mary');
INSERT INTO STUDENT (NAME) VALUES ('Joan');
select
ID
,NAME
,CONCAT(
CAST((ABS(CHECKSUM(NEWID()))%10) as varchar(1))
, CHAR(ASCII('a') + (ABS(CHECKSUM(NEWID())) % 25))
, CHAR(ASCII('A') + (ABS(CHECKSUM(NEWID())) % 25))
, LEFT(LOWER(NEWID()),2)
, LEFT(NEWID(),2)
, LEFT(LOWER(NEWID()),2)
) AS PASSWORD
from STUDENT
SQL Fiddle
However, you may wish to Make your own generator taking into account a high level of complexity generated password.
https://www.simple-talk.com/blogs/2009/09/30/strong-password-generator/
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
I have a function which I want to be more efficient. I know that most of the cost is in seeking index. I want to know if there is a way to merge these two index seeks into one.
Here is my function
ALTER FUNCTION [dbo].[Search]
(
#Query nvarchar(150)
)
RETURNS #result TABLE
(
ID int,
ResultPhrase nvarchar(max),
Domain nvarchar(50)
)
AS
BEGIN
declare #id int,
#OffsetArticle nvarchar(50),
#OffsetProfile nvarchar(50),
#index int,
#domain nvarchar(50),
#StringResult nvarchar(max),
#substrIndex int,
#substrCount int,
#field nvarchar(40),
#begin int,
#end int
DECLARE pidCursor CURSOR FAST_FORWARD
FOR select id,OffsetArticle,OffsetProfile from PerfTest where contains (data,#Query)
open pidCursor
FETCH NEXT FROM pidCursor INTO #id,#OffsetArticle,#OffsetProfile
while (##FETCH_STATUS = 0)
begin
select #index = CHARINDEX(#Query,data) from PerfTest where ID=#id
exec #domain = dbo.FindDomain #id=#id ,#index=#index, #offsetArticle=#OffsetArticle, #offsetProfile=#OffsetProfile
if(#domain = N'Profile')
set #end = SUBSTRING(#offsetProfile,CHARINDEX(N'-',#offsetProfile)+1,len(#offsetProfile)-CHARINDEX(N'-',#offsetProfile))
set #begin = SUBSTRING(#offsetProfile,1,CHARINDEX(N'-',#offsetProfile)-1)
if(#domain = N'Article')
set #end = SUBSTRING(#OffsetArticle,CHARINDEX(N'-',#OffsetArticle)+1,len(#OffsetArticle)-CHARINDEX(N'-',#OffsetArticle))
set #begin = SUBSTRING(#OffsetArticle,1,CHARINDEX(N'-',#OffsetArticle)-1)
if(#index - 20 < CAST(#begin as int))
set #substrIndex = #index
else
set #substrIndex = #index - 20
if(#index + 150 > CAST(#end as int))
set #substrCount = #end
else
set #substrCount = #index + 150
select #StringResult = SUBSTRING(data,#substrIndex,#substrCount) from PerfTest where ID=#id
insert #result (id,ResultPhrase,Domain) values (#id,#StringResult,#domain)
if(##FETCH_STATUS = 0)
FETCH NEXT FROM pidCursor INTO #id,#OffsetArticle,#OffsetProfile
end
CLOSE pidCursor
DEALlOCATE pidCursor
RETURN
END
For combining the #index and #string result query, you could do the following:
DECLARE #results TABLE
(index INT, stringresult NVARCHAR(MAX))
INSERT INTO #results(Index,stringresult)
SELECT CHARINDEX(#Query,data)
,SUBSTRING(data,#substrIndex,#substrCount
FROM PerfTest where ID=#id
SET #index = (SELECT index from #results)
SET #stringresults = (SELECT stringresult FROM #results)
This will read the perftest table only once, then read the results from a in memory table into the two variables
Include
CHARINDEX(#Query,data)
in the cursor, instead of calculating it first thing inside the loop. That should cut of about half of the seeks. While seeks are the most efficient access way for SQL Server, eliminating half of them is still going to help any query.
You could also try to include data in the cursor and calculate #StringResult using that value. If that is going to help is very dependent on the avg size of the data column. If it is small enough that should speed things significantly. What "small enough" is is dependent on a lot of things and you should try it out. If data is VARCHAR(1000) pulling it in the cursor it is most likely going to help, however if it is VARCHAR(MAX) with and average length of 2 MB it is most likely going to hurt.
Finally, as Pete mentioned, the best solution would be to write this in a set-based fashion, but that might be more complex, as you can't call a stored procedure from inside a single select statement.
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.