Been wracking my brains with this

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Been wracking my brains with this

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SQLServerMS SSC Veteran Points: 243 More actions · Answer 1

Hi Judo,

Execute the below query, you will be getting the result as you said

CREATE TABLE #TIMESHEET(TimesheetKey INT, SiteKey INT, Date VARCHAR(10), Duration INT, GroupRule INT)

INSERT INTO #TIMESHEET VALUES(1000, 1, '7/28/2011', 480, 1)

INSERT INTO #TIMESHEET VALUES(1001, 1, '7/28/2011', 600, 1)

INSERT INTO #TIMESHEET VALUES(1002, 1, '7/28/2011', 480, 2)

SELECT * FROM #TIMESHEET

SELECT

B.[TIMESHEETKEY]

, B.[SITEKEY]

, B.[DATE]

, A.[SUM_DURATION]

, A.[GROUPRULE]

FROM

(

SELECT

[GroupRule]

, SUM([DURATION]) [SUM_DURATION] FROM #TIMESHEET

GROUP BY GroupRule

) A

JOIN #TIMESHEET B

ON A.[GroupRule] = B.[GroupRule]

Have a nice day!

ChrisM@Work SSC Guru Points: 186127 More actions · Answer 2

SELECT TimesheetKey, SiteKey, [Date],

SUMDuration = SUM(Duration) OVER(PARTITION BY SiteKey, [Date], GroupRule), GroupRule

FROM #TIMESHEET

^{“Write the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw}

For fast, accurate and documented assistance in answering your questions, please read this article.
Understanding and using APPLY, (I) and (II) Paul White
Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

SQLServerMS SSC Veteran Points: 243 More actions · Answer 3

SELECT TimesheetKey, SiteKey, [Date],

SUMDuration = SUM(Duration) OVER(PARTITION BY GroupRule), GroupRule

FROM #TIMESHEET

Have a nice day!

judo-467517 Newbie Points: 6 More actions · Answer 4

I forgot to thank you guys. The PARTITION BY worked out greatly!