Jun 10, 2007 #1 Hayton Technical User Oct 17, 2001 257 NZ Hi Guys, =IIf(IsError([txtFR1]/[txtTot]),0,[txtFR1]/[txtTot]) For the formula above, when txtTot is 0, it shows the #Num! error. I have tried using IsError function to get around it and it does not work. Any suggestions would be appreciated. Hayton McGregor
Hi Guys, =IIf(IsError([txtFR1]/[txtTot]),0,[txtFR1]/[txtTot]) For the formula above, when txtTot is 0, it shows the #Num! error. I have tried using IsError function to get around it and it does not work. Any suggestions would be appreciated. Hayton McGregor
Jun 10, 2007 #2 PHV MIS Nov 8, 2002 53,708 FR Why not simply this ? =IIf([txtTot]=0,0,[txtFR1]/[txtTot]) Hope This Helps, PH. FAQ219-2884 FAQ181-2886 Upvote 0 Downvote
Why not simply this ? =IIf([txtTot]=0,0,[txtFR1]/[txtTot]) Hope This Helps, PH. FAQ219-2884 FAQ181-2886
Jun 10, 2007 Thread starter #3 Hayton Technical User Oct 17, 2001 257 NZ Hi PHV, Thanks for the reply. Hayton McGregor Upvote 0 Downvote