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Hello,
I'm trying to take a column containing past dates (in the format: MM/DD/YYYY); compare it to a CONSTANT, more current date (e.g. 04/01/2007); calculate the difference in the number of days (my first choice) or months (second choice); then populate that result in a corresponding column in that same Excel sheet. Your help is always appreciated and thanks in advance! Chris |
#2
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Hi Chris,
Try function days360() -- Pranav Vaidya VBA Developer PN, MH-India If you think my answer is useful, please rate this post as an ANSWER!! "Not Excelerated" wrote: Hello, I'm trying to take a column containing past dates (in the format: MM/DD/YYYY); compare it to a CONSTANT, more current date (e.g. 04/01/2007); calculate the difference in the number of days (my first choice) or months (second choice); then populate that result in a corresponding column in that same Excel sheet. Your help is always appreciated and thanks in advance! Chris |
#3
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Hi,
It's done with subtratction and format the cell as general. For example with your constant date in C1 and your earlier dates in column A in B1 try:- =$c$1-A1 Drag down. Mike "Not Excelerated" wrote: Hello, I'm trying to take a column containing past dates (in the format: MM/DD/YYYY); compare it to a CONSTANT, more current date (e.g. 04/01/2007); calculate the difference in the number of days (my first choice) or months (second choice); then populate that result in a corresponding column in that same Excel sheet. Your help is always appreciated and thanks in advance! Chris |
#4
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For days it's a simple subtraction
=current_date-past_date format result as general for months =DATEDIF(past_date,current_date,"m") -- Regards, Peo Sjoblom "Not Excelerated" wrote in message ... Hello, I'm trying to take a column containing past dates (in the format: MM/DD/YYYY); compare it to a CONSTANT, more current date (e.g. 04/01/2007); calculate the difference in the number of days (my first choice) or months (second choice); then populate that result in a corresponding column in that same Excel sheet. Your help is always appreciated and thanks in advance! Chris |
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