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  1. #1
    Join Date
    Mar 2004
    Location
    California
    Posts
    502

    Unanswered: IIF([Field3]="S" [Field4]=[Field4]/6)

    Seems logical enough for me!

    But I get the word "error" in the report for [field4]/6)

    The error description is "missing operator"

    I'm looking up Operators now.

    Rick

  2. #2
    Join Date
    Jun 2005
    Location
    Richmond, Virginia USA
    Posts
    2,763
    Provided Answers: 19
    First off, you need a separator (comma) after the condition

    IIF([Field3]="S", [Field4]=[Field4]/6)

    Secondly, IIF() takes two arguments! You have the first, or True argument, but you haven't told Access what to do if

    IIF([Field3]<>"S"

    So you need something like

    IIF([Field3]="S", [Field4]=[Field4]/6, [What to do if False goes here])

    Linq ;0)>
    Hope this helps!

    The problem with making anything foolproof...is that fools are so darn ingenious!

    All posts/responses based on Access 2003/2007

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