I need help in inserting a character inside a string, e.g.:
031613 05:39 AM
The output should be:
03/16/13 05:39 AM
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3 Answers
You can use STUFF
DECLARE @String NVARCHAR(20) = '031613 05:39 AM'
SELECT STUFF(STUFF(@String,3,0,'/'),6,0,'/')
Comments
How about using SUBSTRING?
DECLARE @String VARCHAR(50) = '031613 05:39 AM'
SELECT @String,
SUBSTRING(@String,1,2) + '/' + SUBSTRING(@String,3,2) + '/' + SUBSTRING(@String,3,2) + SUBSTRING(@String,7,LEN(@String)-6)
SQLFiddle DEMO
1 Comment
ajdeguzman
Thanks!Can you also use STUFF T-SQL for that problem?
use can also do like this.. but quite big solution.
declare @T varchar(100)= '031613 05:39 AM'
declare @result varchar(100) =''
;with CTE
as
(
Select 0 as startIndex, 2 as endIndex
union all
select convert(int, endIndex) as startIndex , endIndex+2 as endIndex from CTe where endIndex < charindex(' ',@T)
)
select top 3 @result += substring(@T,startIndex+1, endIndex-startIndex)+'/' from CTe
select substring( @result,0,len(@result))
