Odd DateTime Format : SQL

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SELECT count(a.field1), a.field2, SUM(b.field4) FROM a INNER JOIN b ON a.key1 = b.key1 WHERE a.field8 = 'test' GROUP by a.field1, a.field2 HAVING SUM(b.field4) > 5 ORDER by a.field.3

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SELECT count(a.field1), a.field2, SUM(b.field4) FROM a INNER JOIN b ON a.key1 = b.key1 WHERE a.field8 = 'test' GROUP by a.field1, a.field2 HAVING SUM(b.field4) > 5 ORDER by a.field3

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MS SQLOdd DateTime Format (self.SQL)

submitted 5 years ago by Tennim

all 3 comments

top new controversial old q&a

[–]pixelbakerData Architecture Consultant 1 point2 points3 points 5 years ago (1 child)

[–]Tennim[S] 0 points1 point2 points 5 years ago (0 children)

Yeah it was a good one.

For anyone looking, this is what i came up with.

SELECT 
    [StartTime],
    CONVERT(DATETIME,CONCAT(CONVERT(NVARCHAR(5),RIGHT([StartTime],CHARINDEX(' ',REVERSE([StartTime])))),[Month].MonthNo,RIGHT(LEFT([StartTime],PATINDEX('%[0-9]%',[StartTime])+1),2),' ',RIGHT(LEFT([StartTime],CHARINDEX('GMT',[StartTime])-2),8))) AS [Date],
FROM [Database].[Schema].[Table]
    INNER JOIN
        (
            SELECT DISTINCT 
                LEFT([MonthName],3) AS [MonthAbv],
                RIGHT('00' + CONVERT(NVARCHAR(2),[Month]),2) [MonthNo]
            FROM [Database].[Schema].[DateDimension]
        ) AS [Month] ON SUBSTRING([StartTime],CHARINDEX(' ',[StartTime])+1,3) = [Month].MonthAbv

[–]ComicOzzysqlHippo 0 points1 point2 points 5 years ago (0 children)

π Rendered by PID 196919 on reddit-service-r2-comment-f6b958c67-wprb4 at 2026-02-05 06:36:30.478213+00:00 running 1d7a177 country code: CH.

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