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

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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gracie1972
Asked:
gracie1972
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1 Solution
 
Patrick MatthewsCommented:
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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gracie1972Author Commented:
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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andrewssd3Commented:
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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gracie1972Author Commented:
OK, I think I found it, just not sure why it is doing what it is doing.

Thanks
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andrewssd3Commented:
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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