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Hello all,
Just wondering if there is someone out there that might have run into this problem before and would be willing to point me in the correct direction. I am attempting to create a macro that will process information based on the row (s) that the user selects. I am attmpting to allow the used to select any row(s) that he or she might want but am having trouble in determining what has been selected. I would like to allow the user to select any cell(s) on any row(s) and only process the information on those rows. An example of this would be a user could select A1, B3, F3 and all of row 100. From this I would want to process row 1, 3 (but only once even though it was selected twice) and 100. From a little research I think that I can read row numbers into and access them from an array variable in a manner of ARRAY_NAME(counter) = Row_Number - Is this correct as well? Any help that anyone would be willing to offer would be greatly appreciated. |
#2
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hi, Terry !
here is one (possible) approach: Sub A() Dim myCell As Range For Each myCell In Intersect(Range("a:a"), Selection.EntireRow) MsgBox "Processing row: " & myCell.Row Next End Sub hth, hector. __ OP __ Just wondering if there is someone out there that might have run into this problem before and would be willing to point me in the correct direction. I am attempting to create a macro that will process information based on the row (s) that the user selects. I am attmpting to allow the used to select any row(s) that he or she might want but am having trouble in determining what has been selected. I would like to allow the user to select any cell(s) on any row(s) and only process the information on those rows. An example of this would be a user could select A1, B3, F3 and all of row 100. From this I would want to process row 1, 3 (but only once even though it was selected twice) and 100. From a little research I think that I can read row numbers into and access them from an array variable in a manner of ARRAY_NAME(counter) = Row_Number - Is this correct as well? Any help that anyone would be willing to offer would be greatly appreciated. |
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