To trap for an error, you need to do it in the following fashion:

=if(iserr(formula),"",formula)

Next, you need to spell the function properly. It's "networkdays", not
"networkdyas".

Thirdly, your parentheses must match. You have a "9" after networkdyas, rather
than the required opening parenthesis.

Fourth, you need to identify the worksheet properly. Sometimes you have "bldg8"
(without a space), and sometimes "bldg 8" (with a space).

Fix these errors, and the formula should work. Post back if you need more help.
Don't expect us to continually fix typographic errors, though.

--
Regards,
Fred


"Cham" wrote in message
...
Hi,
I have a formula like this:


IF(IF('BLDG8'A62='BLDG 7'!A62,networkdyas9'BLDG 8'!c62,'bLDG 7'!c62),
"")=ISERR('Bldg 8'!d74), 'Bldg 8'!74,"")

the first If function returns a "#Value", my attemp on the above function is
to get a text value answer if the formula comesa out as #VAlue. I .e. the
IF('BLDG8'A62='BLDG 7'!A62,networkdyas9'BLDG 8'!c62,'bLDG 7'!c62)=
#Value,(which is an error), then give the Sheet X Cellx Value as a end value.
the sheet X value is a written statement.

Any suggestions on how to do this? or even improve the IF statement. Because
not all cells will have a text input, tmajority have dates.


--
Cham