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Convert 6 digit letter to month and Year

  • February 27, 2015 at 3:59 pm

    #315866

    Can any one share how to convert 6 digit number to mm/yyyy.

    Thanks in Advance.

  • February 27, 2015 at 4:17 pm

    #1780827

    Abhiram (2/27/2015)

    Can any one share how to convert 6 digit number to mm/yyyy.

    Thanks in Advance.

    What is contained in the 6 digit number?

    --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)

  • February 27, 2015 at 4:44 pm

    #1780832

    I have a column 062008 I want to display that number as 06/2008.

  • February 27, 2015 at 5:17 pm

    #1780835

    Abhiram (2/27/2015)

    I have a column 062008 I want to display that number as 06/2008.

    SELECT STUFF(yourcolumn,3,0,'/')

    --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)

  • February 27, 2015 at 11:51 pm

    #1780856

    Further on Jeff's fine solution, you may want to consider converting the column from (var)character data type to date data type. Not only does it take less storage (3 bytes), it also enables you to do calculation on the column.

    😎

    Extending the previous method to return a date data type

    SELECT CONVERT(DATE,'01/' + STUFF(yourcolumn,3,0,'/'),103)

    Alternative method without any string manipulation

    SELECT DATEADD(MONTH

    ,(((CONVERT(INT,yourcolumn,0) -1) % 10000) * 12)

    + (CONVERT(INT,yourcolumn,0) / 10000) -1

    ,CONVERT(DATE,'01/01/0001',103))

  • February 27, 2015 at 11:54 pm

    #1780857

    Thank you very much. It worked.

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