Attribute Value Has Quotes

You have a couple of options.

1) you could use the substring function, use the length function - 1 to find the end.

2)  either of the functions below could work.  You can put the double quote in single quotes.

USE [impresario]
GO

/****** Object: UserDefinedFunction [dbo].[LF_STRIP_FORMATTING] Script Date: 10/10/2023 3:41:04 PM ******/
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER OFF
GO

CREATE FUNCTION [dbo].[LF_STRIP_FORMATTING] (
@StrIn NVARCHAR(1024)
)
RETURNS NVARCHAR(1024)
AS

/****************************************************************************************/
/* LF_STRIP_FORMATTING */
/* Created By: Katie Lachance */
/* Date Created: March 5, 2007 */
/* */
/* Strips @StrIn of its formatting -- removes spaces and non-alphanumeric characters. */
/****************************************************************************************/

BEGIN

DECLARE @StrOut NVARCHAR(1024)

SET @StrOut =
CASE
WHEN @StrIn = '' THEN NULL
ELSE RTRIM(@strIn)
END

SET @StrOut = REPLACE(@StrOut,' ','')
SET @StrOut = REPLACE(@StrOut,'.','')
SET @StrOut = REPLACE(@StrOut,'-','')
SET @StrOut = REPLACE(@StrOut,'(','')
SET @StrOut = REPLACE(@StrOut,')','')
SET @StrOut = REPLACE(@StrOut,'[','')
SET @StrOut = REPLACE(@StrOut,']','')
SET @StrOut = REPLACE(@StrOut,',','')
SET @StrOut = REPLACE(@StrOut,'/','')
SET @StrOut = REPLACE(@StrOut,'\','')
SET @StrOut = REPLACE(@StrOut,'#','')
SET @StrOut = REPLACE(@StrOut,'''','')

RETURN @StrOut

END

GO

ALTER AUTHORIZATION ON [dbo].[LF_STRIP_FORMATTING] TO SCHEMA OWNER
GO

GRANT EXECUTE ON [dbo].[LF_STRIP_FORMATTING] TO [ImpUsers] AS [dbo]
GO

GRANT EXECUTE ON [dbo].[LF_STRIP_FORMATTING] TO [tessitura_app] AS [dbo]
GO

-----------------------------------------------------------------------------------

USE [impresario]
GO

/****** Object: UserDefinedFunction [dbo].[FS_REPLACE_NON_ALPHANUMERICS] Script Date: 10/10/2023 3:46:04 PM ******/
SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION [dbo].[FS_REPLACE_NON_ALPHANUMERICS] (@in_string VARCHAR(4000), @replace_char VARCHAR(255))
RETURNS VARCHAR(4000)
WITH SCHEMABINDING
AS

/********************************************************************************
New Procedure CWR 5/15/2013 to replace non-alphanumerics with a character
Modified CWR 6/27/2013 #2394 to work with multiple consecutive non-alphanumeric characters

declare @description varchar(30) = 'cont-dt'

Select description = replace(replace(replace(@description, '-', '_'), ':', '_'), '&', '_')

select description = dbo.FS_REPLACE_NON_ALPHANUMERICS(@description, '123')

*********************************************************************************/
BEGIN

DECLARE @searchpat VARCHAR(4000) = '%[^' + @replace_char + ',0-9,A-Z,a-z]%'
DECLARE @out_string VARCHAR(4000)
IF @in_string IS NULL
BEGIN
SELECT @out_string = NULL
RETURN @out_string
END

SET @out_string = @in_string

WHILE PATINDEX(@searchpat, @out_string) > 0
BEGIN
SET @out_string = REPLACE(@out_string, SUBSTRING(@out_string,PATINDEX(@searchpat, @out_string),1),@replace_char)
END

RETURN @out_string
END

GO

ALTER AUTHORIZATION ON [dbo].[FS_REPLACE_NON_ALPHANUMERICS] TO SCHEMA OWNER
GO

GRANT EXECUTE ON [dbo].[FS_REPLACE_NON_ALPHANUMERICS] TO [ImpUsers] AS [dbo]
GO

GRANT EXECUTE ON [dbo].[FS_REPLACE_NON_ALPHANUMERICS] TO [tessitura_app] AS [dbo]
GO

I think I understand.  LOL