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Well, one way would be
addr = Cells(1, 3).Address But one very rarely needs to use A1-style addresses when one has row and column numbers. Using the range object returned by Cells(1, 3) directly is far more efficient than addr = Cells(1, 3).Address Range(addr) ... In article , "Sean Farrow" wrote: Chers both, just what I needed. As a final point onthis: How do I transform an address from all numeric values, e.g f,3 to letter-number form e.g a3. |
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