Convert to datetime

How do I convert this string to datetime?
'25.3.2012 19:41:07'

DECLARE @TimeDay datetime
SET @TimeDay='25.3.2012 19:41:07'
johnkainnAsked:
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chinawalCommented:
DECLARE @TimeDay datetime
SET @TimeDay= replace('25.3.2012 19:41:07','.','/')

Assuming that your server is  dd/mm/yyyy format
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johnkainnAuthor Commented:
When I run this I get following error:
The conversion of a varchar data type to a datetime data type resulted in an out-of-range value.

 When I select GETDATE() I get 2012-03-25 21:30:54.107

When I use following it works
DECLARE @TimeDay datetime
SET @TimeDay='2012-03-25 20:01:54.937'
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chinawalCommented:
oK. try this. this should work.


DECLARE @TimeDay datetime
SET @TimeDay= convert(datetime, '25.3.2012 19:41:07', 104)
select @TimeDay
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waltersnowslinarnoldCommented:
Hi, use the following query sample, this would help you;


DECLARE @v DATETIME
SELECT @v = CONVERT(DATETIME,'25.3.2012 19:41:07',105)
SELECT @v
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Scott PletcherSenior DBACommented:
One of the only 100% safe date formats is YYYYMMDD, so it's safest -- although the code is longer -- to convert to 'YYYYMMDD HH:MM:SS.sss', which works under ALL SQL date and language settings.


DECLARE @TimeString varchar(30)
DECLARE @TimeDay datetime

SET @TimeString = '25.3.2012 19:41:07'

SET @TimeDay = PARSENAME(LEFT(@TimeString, CHARINDEX(' ', @TimeString) - 1), 1) +
RIGHT('0' + PARSENAME(LEFT(@TimeString, CHARINDEX(' ', @TimeString) - 1), 2), 2) +
RIGHT('0' + PARSENAME(LEFT(@TimeString, CHARINDEX(' ', @TimeString) - 1), 3), 2) +
SUBSTRING(@TimeString, CHARINDEX(' ', @TimeString), 30)

SELECT @TimeDay
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Microsoft SQL Server 2008

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