I need help creating an excel spreadsheet where I can have a list of all of my clients and a list of all of the colleges they have applied to and be able to filter by different variables. For example, not only do I need to be able to see the name of the client and all of the colleges they applied to, but I need to also be able to sort by a specific college and see all the clients that have applied to that college. Any help would be greatly appreciated, thank you!
Luna1,
Here's an exammple of how it should be laid out to get the functionality you want:
Client Applied to College Client1 College1 Client1 College2 Client2 College2 Client2 College4 Client2 College5 Client3 College1 Client3 College3 Client3 College4 Client4 College1 Client4 College2 Client4 College3 Client4 College4 Client4 College5
Hope that helps,
~tigeravatar
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Hi tigeravatar,
Thank you for the post. Once I set up my clients and colleges as you have described, how can I easily extract which clients have applied to a particular college? The list of colleges is going to be quite expensive..is there a certain formula I can use?
You would use Excel's autofilter. Here's some links for more information:
http://www.contextures.com/xlautofilter01.html
http://office.microsoft.com/en-us/ex...010073941.aspx
http://office.microsoft.com/en-us/ex...010082314.aspx
Hope that helps,
~tigeravatar
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Ok, great thank you.
One last thing, is there a way to assign multiple colleges to one client without having to list the clients name more than once? It will be difficult to easily see a count of how many clients there are when their name is listed more than once.
Because A1 is a header, the range starts on A2. Adjust the ending cell to suit. You can use this formula to get the number of unique, non-blank entries in the range:
=SUMPRODUCT(--(A2:A10<>"")/COUNTIF(A2:A10,A2:A10&""))
Hope that helps,
~tigeravatar
Forum Rules: How to use code tags, mark a thread solved, and keep yourself out of trouble
A slightly different approach to tigeravtars solution could perhaps be of interest?
Alf
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