Hi All,
What is the cyntax for "IF" statement

I like to test a cell for a value 29.7 and < 30.3

Thanks for the help.
Syed


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saziz
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=if(and(a129.7,a1<30.3),"between","not between")
or
=if(and(29.7<a1,a1<30.3),"between","not between")

I think I like the second one. It kind of looks like A1 should be trapped
between the two values.



saziz wrote:

Hi All,
What is the cyntax for "IF" statement

I like to test a cell for a value 29.7 and < 30.3

Thanks for the help.
Syed

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Dave Peterson

Hi!

Try this:

=IF(AND(A129.7,A1<30.3),"yes","no")

Biff

"saziz" wrote in
message ...

Hi All,
What is the cyntax for "IF" statement

I like to test a cell for a value 29.7 and < 30.3

Thanks for the help.
Syed


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saziz
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You don't tell us what you want to do with the IF statement, but it
would look something like this:

=IF(AND(A129.7,A1<30.3),"In Range","Out of Range")

Does this work for you?

Bruce


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Dave Peterson Wrote:
=if(and(a129.7,a1<30.3),"between","not between")
or
=if(and(29.7<a1,a1<30.3),"between","not between")

I think I like the second one. It kind of looks like A1 should be
trapped
between the two values.

Dave Peterson

Dave is into bondage? All righty, then. It gets the job done.


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Thank you all for help.
Here is the scenario I am trying to fit your cyntax:

If .Value < 30.3 Then
.Offset(0, 1).Value = .Value

if I use this "IF(AND.value(A129.7,A1<30.3)
it gives me an error.

Any idea?
Thanks
syed

Biff Wrote:
Hi!

Try this:

=IF(AND(A129.7,A1<30.3),"yes","no")

Biff

"saziz" wrote in
message ...

Hi All,
What is the cyntax for "IF" statement

I like to test a cell for a value 29.7 and < 30.3

Thanks for the help.
Syed


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Is this a worksheet function or a VBA statement? Your last post makes
me think you are referring to a VBA statement which would look like

If .value29.7 and .value<30.3 then
statement
Else
statement
End if


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MrShorty
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Hmmmm.

http://www.imdb.com/title/tt0119360/



swatsp0p wrote:

You don't tell us what you want to do with the IF statement, but it
would look something like this:

=IF(AND(A129.7,A1<30.3),"In Range","Out of Range")

Does this work for you?

Bruce

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Dave Peterson

Thanks Mr. shorty. That solved my problem.
Syed


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Hi!

Not being unkind, but it helps your helpers if you define the problem
and its context as precisely as you can.

Alf


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