OrientalPearl,
I think you'll find ISNA is an Excel worksheet function. As such VBA does
not understand what it means, hence the compile error.
If you want to use it, you need Application.WorksheetFunctions.ISNA(....

NickHK

"OrientalPearl" wrote in message
ups.com...
Thanks Dave for your prompt help! The two solutions both work well. Re
iserror, I looked up ISNA() and it seems more specialised on working
with '#N/A'. But why doesn't it in this case? Also, '.Text' and
'.Value' do things diffferently? (apparently they make the difference
here)

Ok. I will stick to 'Exit Sub' then. Save a bit of effort in choosing
from the two ;-)

You help is highly appreciated!! Thanks!