I have a report Where my detail row is already summed up from my query. So example on of my fields is called SumOfGeneral_Repair.


Ok so in my report footer i have all my detail rows summed up again. I also have a textbox that is counting the rows that do not have a zero. So what i have in my report footer for formula is:
=Sum([SumOfGeneral_Repair])/[text39]

How would i incorparate that formula so i can forget the divide by zero error?

You can test in an IIf() and only do the division if it's greater than zero.

OK So how would i do that in my formula that i posted?

OK So how would i do that in my formula that i posted?

The Syntax of IIf() is

IIf({Condition to Test}, {Value if Test is True}, {Value if Test is False})

Iif([text39]>0,Sum([SumOfGeneral_Repair])/[text39] ,Sum([SumOfGeneral_Repair])

Disclaimer: I am not sure if this is correct.

Iif([text39]>0,Sum([SumOfGeneral_Repair])/[text39] ,Sum([SumOfGeneral_Repair])

Disclaimer: I am not sure if this is correct.
I think you might want to treat the Null or 0 divisor as having a 0 returned value.

Note: the extra set of parenthesis are for clarification only and can be removed if you desire. I like to use them in formulas like this.

IIf([text39]>0, (Sum([SumOfGeneral_Repair])/[text39]) ,0)

I thionk you might want to treat the Null or 0 divisor as having a 0 returned value.

LOL, I have never been good at math..... :)

LOL, I have never been good at math..... :)


Did that formula work for you?

Did that formula work for you?

Not sure if you were asking me but I didn't test it. My thinking and reasoning skills are a little off today, due to the fact that I am tired..... So my reasoning was wrong when I worked it out in my head.

Not sure if you were asking me but I didn't test it. My thinking and reasoning skills are a little off today, due to the fact that I am tired..... So my reasoning was wrong when I worked it out in my head.


Once you have worked things out, let us know if you have any more questions.

I wouldn't use

IIf([text39]>0,

I would use

IIf([text39]<>0,

Because what if you have a negative number? It would get bypassed.

Use the Nz function. =nz(me.text39/5, 0) if the me.text = 0 then a null value would be the result. with the nz function it will return a 0 in the code I have included.

Use the Nz function. =nz(me.text39/5, 0) if the me.text = 0 then a null value would be the result. with the nz function it will return a 0 in the code I have included.

This will not work. If you have /5 then it will but the problem is if the field is a zero that you are dividing BY.

I beg your pardon, boblarson is correct.

KRYST51, worked good. thanks.