#DIV/0!

[ExcelRoy]

hello,

i have had a problem that you solved for me earlier in the week, i had the #n/a coming up with uncalculated formula in (looked unprofessional)

i used

=IF(ISNA(INDEX(Supplier!$C$5:$C$505,MATCH(B6,Supplier!$A$5:$A$505,0),0)),"",INDEX(Supplier!$C$5:$C$505,MATCH(B6,Supplier!$A$5:$A$505,0),0))

to get round problem

now i have same but with #DIV/0! coming up is there a way round this one

kind regards

 

[Aladin Akyurek]

hello,

i have had a problem that you solved for me earlier in the week, i had the #n/a coming up with uncalculated formula in (looked unprofessional)

i used

=IF(ISNA(INDEX(Supplier!$C$5:$C$505,MATCH(B6,Supplier!$A$5:$A$505,0),0)),"",INDEX(Supplier!$C$5:$C$505,MATCH(B6,Supplier!$A$5:$A$505,0),0))

to get round problem

now i have same but with #DIV/0! coming up is there a way round this one

kind regards

What is the formula that comes up with #DIV/0!?

 

[ExcelRoy]

=SUM(C199*Supplier!I199)

a simple one compared to the last one

thanks

 

[gaj104]

Hi,

Use the iserror function

=IF(ISERROR(C199*Supplier!I199)="TRUE","",(C199*Supplier!I199))

Hope it helps

 

[pgc01]

Hi ExcelRoy

Why the SUM?

You are just multiplying 2 cells.

=C199*Supplier!I199

Please explain
PGC

 

[barry houdini]

=SUM(C199*Supplier!I199)

You wouldn't normally expect to get a #DIV/0! error unless you're dividing by zero. I'm not sure how you get that from that formula unless either C199 or supplier!I199 contains a #DIV/0! error. If so you might want to try to prevent that.

If you have a formula like

=A1/B1 then to prevent #DIV/0! errors change to

=IF(B1,A1/B1,0) or similar

note: in the above formula the SUM is redundant, this formula would suffice

=C199*Supplier!I199

 

[ExcelRoy]

Thanks harry, my formula would read

=IF(G191,F191/G191,0)

thinks shows a value of 0 until i enter the criteria, than it calculates perfectly

But

can you add something to remove the 0, so that it looks more professional

too many 0's showing in a row

 

[Richard Schollar]

Maybe:

=IF(G191,F191/G191,"")

Richard