Re: [libreoffice-users] Issue with INDEX function

The trim function did the trick.  I have been
      burned by this in the past, so don't know why I didn't think of
      it!  It is really the only thing that makes sens.gsOn 08/22/14 13:07, Mark Bourne wrote:<br>
Gregory
      Smith - Oracle Sparc Verification wrote:AllI have created a one dimensional array

        called CATEGORY.When I do this function: =Index(Category,25,0)I

        get the contents       of that index - lets say it is the word "cats"When I do
        this function =if((index(category,25,0)="cats"),1,0)I       get FALSE?!!Anyone have any 
ideas what I am doing
        wrong?gsWith that formula, you should get 0 if the strings don't match (or
      1 if they do); do you really get FALSE displayed in the cell?Assuming you actually get 0 
(because the condition
      (index(category,25,0)="cats") is evaluating to FALSE), make sure
      there are no spaces in the values. "cats " is not equal to "cats",
      even though they look the same. If spurious spaces are likely to
      be a problem, you may want to use:  =if((trim(index(category,25,0))="cats"),1,0)to remove 
spaces either side of the value (and collapse multiple
      spaces within the value into single spaces) before comparison.Mark.
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