I am trying to create a string containing the first character of a string and the first character following each occurrence of a specific character in the string. For example if '.' is the specific character, 'Hello.Bye.Today.Yesterday' should return 'HBTY'.
What I have so far:
SELECT substring(Hello.Bye.Today.Yesterday, 1,1)
If you want to use SQL for this, create a function with the following logic
DECLARE #value VARCHAR(255)
SET #value = 'Hello.Bye.Today.Yesterday'
DECLARE #result VARCHAR(255)
DECLARE #curChar CHAR(1)
DECLARE #start INT, #end INT
SET #start = 2
SET #end = LEN(#value)
SET #result = SUBSTRING(#value, 1,1)
WHILE #start < #end
BEGIN
SET #curChar = SUBSTRING(#value, #start, 1)
IF #curChar = '.'
BEGIN
SET #result = #result + SUBSTRING(#value, #start + 1, 1)
SET #start = #start + 1
END
SET #start = #start + 1
END
Result
HBTY
If your database supports the INSTR() function, this will work. But, as Gordon Linoff pointed out, there are limits to this type of activity in relational databases. As you see, you must hard-wire:
the very first character
each possible occurrence of '.'; the first, the second, the third ...
Not so easy ...
-- use '+' instead of '||' in SQL Server
WITH
input(s) AS (SELECT 'Hello.Bye.Today.Yesterday')
SELECT
SUBSTR(s,1,1)
||SUBSTR(s,INSTR(s,'.',1,1)+1,1)
||SUBSTR(s,INSTR(s,'.',1,2)+1,1)
||SUBSTR(s,INSTR(s,'.',1,3)+1,1)
AS the_string
FROM input;
the_string
HBTY
So you have an exceptional behaviour for the first character to concatenate; then, you have to stop going SUBSTR(s,INSTR(s,'.',1,<n>)+1,1) at the right point in time, so you have to know in advance how often '.' occurs in the string.
Let's see what happens if we go one past the end:
WITH
input(s) AS (SELECT 'Hello.Bye.Today.Yesterday')
SELECT
SUBSTR(s,1,1)
||SUBSTR(s,INSTR(s,'.',1,1)+1,1)
||SUBSTR(s,INSTR(s,'.',1,2)+1,1)
||SUBSTR(s,INSTR(s,'.',1,3)+1,1)
||SUBSTR(s,INSTR(s,'.',1,4)+1,1) -- there is no 4th '.'
||SUBSTR(s,INSTR(s,'.',1,5)+1,1) -- there is no 5th '.'
AS the_string
FROM input;
the_string
HBTYHH
So INSTR() returns 0 , then, and you get the first character of the string each time. You would have to "decorate" your code with CASE WHEN expressions in that case to catch that....
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 need to rip out the digits of a number i'm returning. I have them in a listing like X0006, X0130, and X1030. I need to pull out X's and 0's and return 6,130,1030 . I've tried RTRIM like this. Anyone have any ideas on what I need to do? I'm sure i'm missing something easy.
Thanks
Use replace to convert X and O to empty strings, cast to a integer to remove leading zeros.
Sample code below.
-- Simple number table
create table #nums
( my_id int identity(1,1),
my_text varchar(32)
);
-- Simplte test data
insert into #nums (my_text)
values
('X10X30X'),
('O00O30O');
-- Remove characters & convert to int
SELECT CAST (REPLACE(REPLACE(my_text, 'X', ''), 'O', '') AS INT) as my_number
FROM #nums
Sample output below.
Here is a function you can create that will do exactly what you're asking for and any future get digits needs.
CREATE FUNCTION [dbo].[fn__getDigits]
(
#string VARCHAR(50)
)
RETURNS int
AS
BEGIN
-- declare variables
DECLARE
#digits VARCHAR(50),
#length INT,
#index INT,
#char CHAR(1)
-- initialize variables
SET #digits = ''
SET #length = LEN(#string)
SET #index = 0
-- examine each character
WHILE (#length >= #index)
BEGIN
SET #char = SUBSTRING(#string, #index, 1)
SET #index = #index + 1
-- add to the result if the character is a digit
IF (48 <= ASCII(#char) AND ASCII(#char) <= 57)
BEGIN
SET #digits = #digits + #char
END
END
-- return the result
RETURN cast(#digits as int)
END
GO
select [dbo].[fn__getDigits]('X0006')
select [dbo].[fn__getDigits]('X0130')
select [dbo].[fn__getDigits]('X1030')
I used replace to get rid of the X and just copied/pasted into excel to get the preceding zeros off
I am trying to return the prefix of a VAT number with a SQL function. Because of some changes in these numbers and differences in countries and mistakes in the database, the length of the prefix differs from 0 to 4 characters. So the input of my function is a string, with a prefix of not numeric characters and then some numbers. For example ES012345678, and then i only want to return ES.
I wrote a function for it and it fails, it only returns NULL, even when the input is like the example.
Anyone knows where my mistake is?
here is my SQL code:
ALTER FUNCTION [dbo].[returnPreOfVat]
(
-- Add the parameters for the function here
#VATstring varchar
)
RETURNS varchar
AS
BEGIN
-- Declare the return variable here
DECLARE #Result varchar
DECLARE #char varchar(2)
DECLARE #counter int
SET #counter =1;
SET #char = '';
WHILE (#counter < 5) --check some from the front
BEGIN
SET #char = SUBSTRING(#VATstring, #counter,1); --get next char from front
IF(ISNUMERIC(#char)<>1) -- not numeric
BEGIN
SET #Result = #Result + #char;
END
SET #counter=#counter+1;
END
-- Return the result of the function
RETURN #Result
END
you never initialize the result , Please try:
DECLARE #Result varchar = ''
If recall correctly NULL + str = NULL.
I think a loop is overkill here. I'd combine the LEFT and PATINDEX functions
select LEFT(#VATstring, PATINDEX('%[0-9]%', #VATstring)-1)
as my table rows has data which is starting and ending with non alphabet char such as ".",";","," etc . I need to remove them from the string . How can I do this ? I have done this using like statements there I need to search for each pattern and filter the. Is there any other better way ?
select lrno,free_text ,
(CASE
WHEN free_text LIKE ',%' THEN SUBSTRING (free_text,2,len(free_text)-1)
END) FREE_TEXT_TRIM
from vwLRClass3_FreeText
ORDER BY PROPERTY_TEXT_VALUE
The Below function will remove other non alphabet chars from a string.
IF object_id('dbo.GetAlphachars') is not NULL
DROP FUNCTION dbo.GetNumeric
GO
CREATE FUNCTION dbo.GetAlphachars (#strAlphachars VARCHAR(256))
RETURNS VARCHAR(256)
AS
BEGIN
DECLARE #intAlpha INT
SET #intAlpha = PATINDEX('%[^a-z]%', #strAlphachars )
BEGIN
WHILE #intAlpha > 0
BEGIN
SET #strAlphachars = STUFF(#strAlphachars, #intAlpha, 1, '' )
SET #intAlpha = PATINDEX('%[^a-z]%', #strAlphachars )
END
END
RETURN ISNULL(#strAlphachars ,0)
END
GO
Example.
CREATE table vish(val varchar(20));
INSERT into vish values('.,,.?ishhh>...,');
SELECT dbo.getalphachars(val) from vish;
result
------------
ishhh