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search a range and return a value from another range without using arrays

  1. #1
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    search a range and return a value from another range without using arrays

    i'm looking for some help with this vba function. I want to make a lookup function that compares the value in a searchfield to the value of a range and returns a partial match.
    for example, if I have a range that contains the word october it would return that word if I search for oct

    The function I have works. However it is pretty slow when applied to large ranges. I've converted all my ranges to arrays and used those to search. I think solving the function without arrays should speed it up considerably.
    can anyone help me with the following two problems:

    * How can I change this function to work without having to convert my ranges to arrays first
    * How do I get this function to work with an entire column


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    Last edited by marcwagner; 12-14-2009 at 05:53 AM. Reason: conforming to mod request

  2. #2
    Forum Expert royUK's Avatar
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    Re: Help me tweak this formula

    Your post does not comply with Rule 1 of our Forum RULES. Your post title should accurately and concisely describe your problem, not your anticipated solution. Use terms appropriate to a Google search. Poor thread titles, like Please Help, Urgent, Need Help, Formula Problem, Code Problem, and Need Advice will be addressed according to the OP's experience in the forum: If you have less than 10 posts, expect (and respond to) a request to change your thread title. If you have 10 or more posts, expect your post to be locked, so you can start a new thread with an appropriate title.
    To change a Title on your post, click EDIT then Go Advanced and change your title, if 2 days have passed ask a moderator to do it for you.
    Hope that helps.

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  3. #3
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    re: search a range and return a value from another range without using arrays

    Hi,

    No need to re-invent the wheel, use the Find method (Range object).
    For a partial match set the LookAt argument to xlPart.

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