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Nested If Find

Posted on 2011-02-23
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Last Modified: 2012-08-14
is it possible to do a nested if find or search on a cell ?

e.g =IF(FIND("Ross",A1),"TRUE",IF(FIND("Turner",A1),"True","False")
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Question by:Ross Turner
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sjklein42 earned 250 total points
ID: 34961766
FIND does not return True/False, you need to use ISERROR to check whether it succeeded or not:

=IF(NOT(ISERROR(FIND("Ross",A1))),"TRUE",(IF(NOT(ISERROR(FIND("Turner",A1))),"TRUE","FALSE")))

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by:Runrigger
ID: 34961768
=IF(AND(ISERROR(SEARCH("Ross",A1)),ISERROR(SEARCH("Turner",A1))),"FALSE","TRUE")

for not case-sensitive

=IF(AND(ISERROR(FIND("Ross",A1)),ISERROR(FIND("Turner",A1))),"FALSE","TRUE")

for case-sensitive
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by:Runrigger
ID: 34961778
=IF(AND(ISERROR(SEARCH("Ross",A1)),ISERROR(SEARCH("Turner",A1))),"FALSE","TRUE")

for not case-sensitive

=IF(AND(ISERROR(FIND("Ross",A1)),ISERROR(FIND("Turner",A1))),"FALSE","TRUE")

for case-sensitive
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by:barry houdini
ID: 34961855
If you want TRUE if either "Ross" or "Turner" is in A1 try

=COUNT(SEARCH({"Ross","Turner",A1))>0

You can put that in an IF if you want e.g.

=IF(COUNT(SEARCH({"Ross","Turner",A1))>0,"x","y")

regards, barry
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by:barry houdini
ID: 34961892
Sorry, I missed out a }, try like this

=COUNT(SEARCH({"Ross","Turner"},A1))>0

or the IF like this

=IF(COUNT(SEARCH({"Ross","Turner"},A1))>0,"x","y")

regards, barry



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by:Runrigger
ID: 34961910
Good one Barry, why doesn't this "=COUNT(SEARCH({"Ross","Turner"},A1))" throw out an error when the values are not found, whereas "=SEARCH({"Ross","Turner"},A1)" does?
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by:barry houdini
ID: 34962035
Hello Runrigger

COUNT just ignores errors so if Ross is found but Turner isn't the SEARCH function returns an array something like this

{10,#VALUE!}

COUNT just counts numbers so in that case the result of the COUNT function is 1, obviously any number > 0 indicates a match with a least one of the "names". I like this approach because you can easily extend it to count many names, you can even use a cell range that contains the names in place of {"Ross","Turner"}.....but that will make the formula into an "array"......

regards, barry
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by:Runrigger
ID: 34962056
ah, ah, got it. I should have realised, COUNT is not a function that I use very often!
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Author Closing Comment

by:Ross Turner
ID: 34962490
That worked beautifully !!!!
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