DATEDIFF with days and hours

  • Hi guys, I am trying to calculate the diffs between to days as a decimal. The hours will an easier wayybe in decimal.

    i can do it in varchar, but i need to be able to agrigate my value.

    eg.

    datediff(DD, [CreatedOn],getdate()) + CAST(DATEPART(HH, GETDATE())- DATEPART(HH, [CreatedOn]) AS REAL)/10

    this gives me the results i need.

    is there an easier way?

    Ian Cockcroft
    MCITP BI Specialist

  • Ian C0ckcroft (2/18/2013)


    Hi guys, I am trying to calculate the diffs between to days as a decimal. The hours will an easier wayybe in decimal.

    i can do it in varchar, but i need to be able to agrigate my value.

    eg.

    datediff(DD, [CreatedOn],getdate()) + CAST(DATEPART(HH, GETDATE())- DATEPART(HH, [CreatedOn]) AS REAL)/10

    this gives me the results i need.

    is there an easier way?

    Why are you dividing the difference in hours by 10? I'm guessing that's supposed to be 24 with a conversation to a float or decimal type.

    Either way I think this is what you are looking for

    Declare @create_date datetime = '1/1/2013', @execution_date DATETIME = GETDATE();

    SELECT CAST(@execution_date - @create_date AS REAL)

  • To get the hours as a decimal. else its 6 days + 5 hours = 11 days and should be 6 days + .5 hours = 6.5 days.

    Not 100% sure this will work yet.

    Ian Cockcroft
    MCITP BI Specialist

  • This should help you

    DECLARE@table TABLE

    (

    IDINT,

    StartDateDATETIME,

    EndDateDATETIME

    )

    INSERT@table( ID, StartDate, EndDate )

    SELECT1, '2013-02-17 10:33:10', '2013-02-17 20:14:40' UNION ALL

    SELECT1, '2013-02-13 12:42:55', '2013-02-14 14:30:50' UNION ALL

    SELECT1, '2013-02-12 15:04:32', '2013-02-15 12:22:25' UNION ALL

    SELECT1, '2013-02-16 20:08:18', '2013-02-18 02:10:10'

    SELECTT.ID, T.StartDate, T.EndDate,

    CASE

    WHEN DATEADD( DAY, DATEDIFF( DAY, T.StartDate, T.EndDate ), T.StartDate ) <= T.EndDate

    THEN DATEDIFF( DAY, T.StartDate, T.EndDate )

    ELSE DATEDIFF( DAY, T.StartDate, T.EndDate ) - 1

    END + --====================Gives you the days

    (

    CASE

    WHEN DATEADD( MINUTE, DATEDIFF( MINUTE, T.StartDate, T.EndDate ), T.StartDate ) <= T.EndDate

    THEN DATEDIFF( MINUTE, T.StartDate, T.EndDate )

    ELSE DATEDIFF( MINUTE, T.StartDate, T.EndDate ) - 1

    END % ( 60 * 24 )

    ) / 1440.00 AS Result --===============Gives you the hours in decimals

    FROM@table AS T

    Edit: Added some comments and changed the query( "<" condition changed to "<=" )


    Kingston Dhasian

    How to post data/code on a forum to get the best help - Jeff Moden
    http://www.sqlservercentral.com/articles/Best+Practices/61537/

  • Ian C0ckcroft (2/18/2013)


    To get the hours as a decimal. else its 6 days + 5 hours = 11 days and should be 6 days + .5 hours = 6.5 days.

    Not 100% sure this will work yet.

    Unless you know of a timekeeping system I am unfamiliar with, 6 days + 5 hours is not the same as 6.5 days 🙂

  • Thanks Kingston, looks much neater, will try that.

    Gabriel, its just 6+5 =11 and should be 6 + .5 = 6.5 🙂

    Ian Cockcroft
    MCITP BI Specialist

  • Ian C0ckcroft (2/18/2013)


    Thanks Kingston, looks much neater, will try that.

    Gabriel, its just 6+5 =11 and should be 6 + .5 = 6.5 🙂

    I'm sorry I did not make the point clear. I am concerned that your math in your formula is incorrect and if you are using this math to test your code against, your debugging is going to be off. 6 days + 5 hours doesn't equal 6.5 days, it equals 6.208 days.

    Please see below:

    6 days + 1 hour = 6.041 days

    6 days + 2 hour = 6.083 days

    6 days + 3 hour = 6.125 days

    6 days + 4 hour = 6.166 days

    6 days + 5 hour = 6.208 days

    6 days + 6 hour = 6.25 days

    6 days + 7 hour = 6.291 days

    6 days + 8 hour = 6.333 days

    6 days + 9 hour = 6.375 days

    6 days + 10 hour = 6.416 days

    6 days + 11 hour = 6.458 days

    6 days + 12 hour = 6.5 days

    6 days + 13 hour = 6.541 days

    6 days + 14 hour = 6.583 days

    6 days + 15 hour = 6.625 days

    6 days + 16 hour = 6.666 days

    6 days + 17 hour = 6.708 days

    6 days + 18 hour = 6.75 days

    6 days + 19 hour = 6.791 days

    6 days + 20 hour = 6.833 days

    6 days + 21 hour = 6.875 days

    6 days + 22 hour = 6.916 days

    6 days + 23 hour = 6.958 days

    6 days + 24 hour = 7 days

  • Gabriel P (2/18/2013)


    Ian C0ckcroft (2/18/2013)


    Hi guys, I am trying to calculate the diffs between to days as a decimal. The hours will an easier wayybe in decimal.

    i can do it in varchar, but i need to be able to agrigate my value.

    eg.

    datediff(DD, [CreatedOn],getdate()) + CAST(DATEPART(HH, GETDATE())- DATEPART(HH, [CreatedOn]) AS REAL)/10

    this gives me the results i need.

    is there an easier way?

    Why are you dividing the difference in hours by 10? I'm guessing that's supposed to be 24 with a conversation to a float or decimal type.

    Either way I think this is what you are looking for

    Declare @create_date datetime = '1/1/2013', @execution_date DATETIME = GETDATE();

    SELECT CAST(@execution_date - @create_date AS REAL)

    Nice to see a kindred spirit. I normally use FLOAT for the same thing but that's basically the way I do it. Nice and simple.

    I wish they had made such simple calculations possible with the new date and time datatypes instead of trying to follow some bloody ANSI/ISO standard for the sake of the myth known as "portability".

    --Jeff Moden


    RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
    First step towards the paradigm shift of writing Set Based code:
    ________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.

    Change is inevitable... Change for the better is not.


    Helpful Links:
    How to post code problems
    How to Post Performance Problems
    Create a Tally Function (fnTally)

  • Thanks Gabriel, see what you mean. I did think about that. Not sure how I'm going to handle that.

    These time diffs are going into a dimension and the reported on. Anyone looking at the report will wonder what 6.5 or 6.2 is. may be I should keep the fields seperate so it clearer, 6 days and 5 hours, or 6.2 days.

    But that has other implications.

    Thanks Jeff, will try that as well

    Ian Cockcroft
    MCITP BI Specialist

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