Re: Wildcard character in an "If" statement?
- From: "T. Valko" <biffinpitt@xxxxxxxxxxx>
- Date: Thu, 22 Feb 2007 17:43:47 -0500
Here's one way:
=IF(COUNTIF(A1,"*stringname*"),B1*0.5,B1*0.75)
Or
C1 = stringname
=IF(COUNTIF(A1,"*"&C1&"*"),B1*0.5,B1*0.75)
Biff
SteveDB1" <SteveDB1@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:44655D9E-0204-429F-8C5B-78EB45F2EB65@xxxxxxxxxxxxxxxx
I want to do what seems to be a similar IF function.
Here is my version.
=IF(a1="*stringname*",b1*1/2,b1*3/4)
My goal is to look in a cell which contains a phrase. The phrase itself
varies {the cell contents could be a name of a person, with (word)
following
it}, but the particular component that I'm seeking either shows up as
(word),
or as (word1). Eg., cell contents being within the dbl quote marks: "Dave
Johnson (word)", or "Danny Thomas (word1)"
Where "word" could be anything.
I've tried already, and it <always> returns a false value-- b1*3/4.
1- can I do this?
2- what would I need to do in order to make it work?
I've also tried the tilda, and question mark. Neither of those are
working.
If however, I just have it look in a cell with a single character, the
equation works fine [=if(a1="c",b1*1/2,b1*3/4)]. I'd assume that it'd work
well too if I had just a single word in the cell, as opposed to a number
of
them. For some reason I just can't get it to work with longer elements,
where
I want to locate a single word within a string of 5 or six words.
If I'm unable to do this, what variation would I need to accomplish this?
Thank you.
.
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