[sql-server] 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?

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

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

i think this solution is faster and simple. i use always CTE/recursive beacuse WHILE is so slow on mssql. 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

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:

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 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

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

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.

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

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).

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.