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  1. #1
    stryder09 is offline Advanced Beginner
    Windows XP Access 2007
    Join Date
    Feb 2011
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    85

    A new equation problem

    Okay I have to build an equation that if it is between 3 different variables to return a certain demerit point.

    If [Top/BottomWTSplitBisectedMeasured]>0.48 but <0.5201 Then return a value of 0
    If [Top/BottomWTSplitBisectedMeasured]>0.47 but <0.5301 Then return a value of 2
    If [Top/BottomWTSplitBisectedMeasured]>0.45 but <0.55 Then return a value of 5
    If [Top/BottomWTSplitBisectedMeasured]>0.55 but <0.45 Then return a value of 20.

    Any help would be appreciated thank you.

    Brad

  2. #2
    pbaldy's Avatar
    pbaldy is offline Who is John Galt?
    Windows XP Access 2007
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    Feb 2010
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    22,521
    I'd love to see the number that is ">0.55 but <0.45".

    Take a look at the Switch() function, with tests like

    [Top/BottomWTSplitBisectedMeasured]>0.48 AND [Top/BottomWTSplitBisectedMeasured] <0.5201
    Paul (wino moderator)
    MS Access MVP 2007-2019
    www.BaldyWeb.com

  3. #3
    stryder09 is offline Advanced Beginner
    Windows XP Access 2007
    Join Date
    Feb 2011
    Posts
    85
    Perfect thank you the following worked for me.

    =IIf([Top/BottomWTSplitBisectedMeasured]>0.48 And [Top/BottomWTSplitBisectedMeasured]<0.5201,0,IIf([Top/BottomWTSplitBisectedMeasured]>0.47 And [Top/BottomWTSplitBisectedMeasured]<0.5301,2,IIf([Top/BottomWTSplitBisectedMeasured]>0.449 And [Top/BottomWTSplitBisectedMeasured]<0.551,5,20)))

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