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Using =IF(find_input_value(J$2,$A199)="",I199,find_input_value(J$2,$A199))

Posted on 2011-09-02
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Last Modified: 2012-05-12
I am working on cracking an old spreadsheet.
I dont understand this formula.

Column J2 = July2011
Column A199 = LiquidationDelin[]FC
Column 199= 73.3%

Can someone explain this?
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Question by:gracie1972
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5 Comments
 
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Accepted Solution

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Patrick Matthews earned 2000 total points
ID: 36476045
find_input_value must be a user defined function.

Go to the VB Editor (Alt+F11 from Excel), and see if the workbook has any code modules.
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Author Comment

by:gracie1972
ID: 36476061
Here is the code associated:

Function find_input_value(target_date, target_row)

    Column = search_array(target_date, array_name, "Column", 2, 45)
   
    'MsgBox array_name(7, 1)
    'MsgBox (Column)

    'Row = search_array(target_row, array_name, "Column", 4, 400)
       
    Row = search_array(target_row, array_name, "Row", 1, 450)
    'MsgBox (Row)
   
    find_input_value = array_name(Row, Column)
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LVL 17

Expert Comment

by:andrewssd3
ID: 36476062
find_input_value looks like a user defined function.  If it's not giving you a #NAME error, it is either defined in the VBA project of this workbook, or an add-in that you have loaded.

Have a look in VBA and see if you can find it. It will be looking for some string in J2 I'm guessing.
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Author Comment

by:gracie1972
ID: 36476144
OK, I think I found it, just not sure why it is doing what it is doing.

Thanks
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LVL 17

Expert Comment

by:andrewssd3
ID: 36521276
Sorry for the delay in responding - we need some more user-defined functions, because search_array and array_name must also be defined in your macro project.  It would be probably simpler if you could post the workbook in question to avoid having to come back to you again.
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