CODEWITHSUNDEEP
sqlsql-servert-sqlsubstringpatindex
Devora
Devora

Reputation: 357

selecting the correct letter from a string

CREATE TABLE #tmpTbl (m VARCHAR(100))  
INSERT INTO #tmpTbl VALUES
 (',[Undergraduate1]')    
,(',[Undergraduate10]')   
,(',[Undergraduate11]')   
;
GO

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
         SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m)) as b from #tmpTbl


drop table #tmpTbl

Hi given the above tmptable and select statement, the result will be as follow.

           a            |      b
-------------------------------------------------------
    ,[Undergraduate     |      1]
    ,[Undergraduate     |      10]
    ,[Undergraduate     |      11]

However i want it to be like this.

           a            |      b
-------------------------------------------------------
    ,[Undergraduate     |      1
    ,[Undergraduate     |      10
    ,[Undergraduate     |      11

How can i achieve that? i tried alot of combination with PATINDEX, LEFT, RIGHT, SUBSTRING,LEN. but cant get right of the ] in column B

Upvotes: 12

Views: 306

Answers (4)

Dave Barker
Dave Barker

Reputation: 6437

Here is an alternative approach that will strip any text and just leave the numbers behind.

SELECT LEFT(subsrt, PATINDEX('%[^0-9]%', subsrt + 't') - 1) 
FROM (
    SELECT subsrt = SUBSTRING(m, pos, LEN(m))
    FROM (
        SELECT m, pos = PATINDEX('%[0-9]%', m)
        FROM #tmpTbl
    ) d
) t

Upvotes: 2

Nick Kavadias
Nick Kavadias

Reputation: 7338

you can use replace to remove the ]. Dodgy, but it achieves what you want

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
     REPLACE(SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m)),']','') as b from #tmpTbl

alternative: reverse the string, substring to remove 1st char, reverse back

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
         REVERSE(SUBSTRING(REVERSE(SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m))),2,LEN(M))) as b from #tmpTbl

Upvotes: 8

Gordon Linoff
Gordon Linoff

Reputation: 1269773

I'm inclined to use stuff() for this purpose:

select replace(stuff(m, 1, patindex(m, '%[0-9]%'), ''), ']', '')

Upvotes: 3

sqluser
sqluser

Reputation: 5672

You can use REPLACE to replace ] with ''

select LEFT(m, PATINDEX('%[0-9]%', m)-1) as a,
         REPLACE(SUBSTRING(m, PATINDEX('%[0-9]%', m), LEN(m)), ']', '') as b from #tmpTbl

Upvotes: 3

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