March 16, 2015 at 4:12 pm
Can Someone please help me in padding leading zeroes for the months in End date
Example: actual data is like 6222007
,11301998 in end date column the following query works fine for 11301998 and converts it as 19981130 which was correct.
But 6222007
fails because month has no leading zero and it converts it as 0076222 which is wrong.
How can i make it as 20070622 with the following code
select (case when replace (ltrim(rtrim(ltrim([end date]))), '|', '') in ('99999999','00000000') then NULL
else substring ([END DATE],5,4)+SUBSTRING([END DATE],1,2)+SUBSTRING([END DATE],3,2) end) as ConvEnd_date
from Mydatabase.dbo.[AccountTable]
Thanks
March 16, 2015 at 5:11 pm
SQListic (3/16/2015)
Can Someone please help me in padding leading zeroes for the months in End dateExample: actual data is like 6222007
,11301998 in end date column the following query works fine for 11301998 and converts it as 19981130 which was correct.
But 6222007
fails because month has no leading zero and it converts it as 0076222 which is wrong.
How can i make it as 20070622 with the following code
select (case when replace (ltrim(rtrim(ltrim([end date]))), '|', '') in ('99999999','00000000') then NULL
else substring ([END DATE],5,4)+SUBSTRING([END DATE],1,2)+SUBSTRING([END DATE],3,2) end) as ConvEnd_date
from Mydatabase.dbo.[AccountTable]
Thanks
select (case when replace (ltrim(rtrim(ltrim(x.[New End Date]))), '|', '') in ('99999999','00000000') then NULL
else substring ([New End Date],5,4)+SUBSTRING([New End Date],1,2)+SUBSTRING([New End Date],3,2) end) as ConvEnd_date
from Mydatabase.dbo.[AccountTable]
cross apply (
select right('0'+RTRIM([END DATE]),8) as [New End Date]
) x
Then look at storing dates as dates, so you don't have this mess - if you can make that change.
MM
select geometry::STGeomFromWKB(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
March 16, 2015 at 5:24 pm
Thank u so much
I also did it with RIGHT FUNCTION
SELECT substring(RIGHT('00000000' + CAST([END DATE] AS VARCHAR), 8),5,4)+substring(RIGHT('00000000' + CAST([END DATE] AS VARCHAR), 8),1,2)+
substring(RIGHT('00000000' + CAST([END DATE] AS VARCHAR), 8),3,2) from
March 16, 2015 at 7:06 pm
SQListic (3/16/2015)
Thank u so muchI also did it with RIGHT FUNCTION
SELECT substring(RIGHT('00000000' + CAST([END DATE] AS VARCHAR), 8),5,4)+substring(RIGHT('00000000' + CAST([END DATE] AS VARCHAR), 8),1,2)+
substring(RIGHT('00000000' + CAST([END DATE] AS VARCHAR), 8),3,2) from
Do you ever have any dates that show up as only 6 digits (mdyyyy) or just 4 digits (mdyy)? Also, what is the datatype of the [End Date] column?
--Jeff Moden
Change is inevitable... Change for the better is not.
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