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Default Question relating to the OR function

Prove what?

I'm sure you know how to evaluate formulas!
--

Regards,

RD
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"Tyro" wrote in message
...
Prove it

Tyro

"Ragdyer" wrote in message
...
<<<"It evaluates "ALL" conditions in IF, AND, OR etc then makes a
choice."

You should not include If() in your blanket statement.

=If(a1=1,"OK",if(a1=2,"NG",If(a1=2/0,"Neither")))

Try 1 or 2 in A1 and you'll see that If() *stops* evaluating at the first
True it finds.

Also, your "etc." is too encompassing and needs elaboration!

For example, Lookup() goes right by those #Div/0! errors, and completes
the
formula calculations:

=Lookup(2,1/(A2:J2<""),A2:J2)
Leave some blank cells in the A2:J2 range, and Lookup() will bypass those
#Div/0! errors that are in the lookup vector.
--
Regards,

RD

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"Tyro" wrote in message
t...
I explained to you earlier, Excel is not a "sophisticated" programming
language. It evaluates "ALL" conditions in IF, AND, OR etc then makes a
choice.
You, in your human mind might terminate the evaluation process earlier,

but
Excel does not. Learn to live with it until Bill Gates the Chairman and
Chief Software Architect at Microsoft learns the basics of logic.

Tyro


"Tyro" wroten message
t...
Perhaps =ISERROR(OR(cond_1,cond-2,....cond_N))
This returns TRUE if any of the OR conditions is an error, FALSE if
not.

Tyrp

"FiluDlidu" wrote in message
...
Now that you made me think about it...

=OR(if(iserror([test1]),0,[test1],if(iserror([test2]),0,[test2],...)

That solved my problem, even though not in a very elegant way...

Cheers.

"Tyro" wrote:

=OR(TRUE,FALSE,FALSE,FALSE) evaluates as TRUE;
=OR(TRUE,FALSE,FALSE,#ERROR)
evaluates as #ERROR. Perhaps you want to enclose your formula in an
ISERROR
function (pre-Excel 2007) or IFERROR (Excel 2007)









 
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