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Find nth instance of a character in a string

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    Find nth instance of a character in a string

    Mod. edit: BrisbaneBob - please don't post q's in another member's thread (no matter how old it is) - post moved to dedicated thread.

    This formula was offered as a method of finding the nth instance of "/" in a string. I have used it successfully. I have run it through Formula Evaluate to see how it works but I still can't figure it out. Can someone be kind enough to explain to me how it works?

    =RIGHT(A1,LEN(A1)-FIND("^^",SUBSTITUTE(A1,"/","^^",LEN(A1)-LEN(SUBSTITUTE(A1,"/","")))))

    Thanks and regards
    Last edited by DonkeyOte; 12-26-2011 at 07:03 PM.

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    Re: Find nth instance of a character in a string

    In answer to the question....

    The formula substitutes the nth instance of "/" with ^^ and then uses ^^ to determine starting point of string to return.

    n is determined by LEN of string less LEN of string having had all instances of "/" removed

    Consider:

    A1 = "apples/pears/oranges"
    Len of A1 is 20

    Determine n
    n = 20-18
    18 being the LEN of A1 if you take out the "/"

    We use n [2] as final argument within SUBSTITUTE call to determine which instance of "/" we wish to replace with "^^"
    (when this argument is excluded [optional] all instances are replaced)

    SUBSTITUTE(A1,"/","^^",2)

    At which point A1 equates to: "apples/pears^^oranges"

    To return "oranges" we return right n characters where n is determined by Len of string [20] less position of ^^ within revised string.

    There are alternatives to the above but it's a fairly common approach.

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