PATINDEX IN SQL NOT WORK WITH SPECIAL CHARACTER

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SQL-Server-2005

it does not work if there is sql character like '%' in input string so what is the solution.
i used replace but its not feasible solution as how many character replaced is a question.

declare @phone varchar(50)
set @phone='212%@@f'

select  REPLACE(@phone
               , SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1)
              , '')

Posted 20-Mar-14 17:57pm

vishal_h

Updated 12-May-16 21:39pm

v2

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vishal_h · Answer 1 · 2014-03-20T18:36:00

Solution 1

SQL server 2008 now its work as i write code but when there is single cote in input it fails

SQL

WHILE(PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone) ) > 0
   BEGIN
   --then remove that one character, then continue
   SET @phone = REPLACE(@phone
   , SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1)
   , '')
   END

Posted 20-Mar-14 18:36pm

vishal_h

Comments

vishal_h 21-Mar-14 0:40am

it fails when input is
declare @phone varchar(50)
set @phone='sdf%d'ff$%'

Rocker-Star 16-May-16 17:14pm

you have to escape single quote (') by adding one more single quote ('') in sql server as it is one of its literals.
try the following
declare @phone varchar(50)
set @phone='sdf%d''ff$%'

hiren soni · Answer 2 · 2016-05-12T21:39:00

I will slightly modify the first answer..

declare @phone varchar(50)
set @phone='212%@@f'

WHILE(PATINDEX('%[^0-9]%', @phone) ) > 0
   BEGIN
   --then remove that one character, then continue
   SET @phone = REPLACE(@phone
   , SUBSTRING(@phone, PATINDEX('%[^0-9]%', @phone), 1)
   , '')
   END

SELECT @phone