Need help with IIF

[twisted]

Hey All,




I have a IIF statement that keeps returning errors. Can I please get someone to point me in the right direction.
I have a excel IF statement I need to pretty much copy in to access query

Excel
=IFERROR(IF(P30>(-15.55*LN(O30)+75.57),1,IF(P30>(-8.14*LN(O30)+39.991),2,IF(P30>(-5.061*LN(O30)+20.856),3,IF(P30>(-2.749*LN(O30)+10.05),4,5)))),"")
P30 is Support
O30 is Stability
Access
Probability: IIf([Support]>(-15.55*Log([Stability])+75.55,"1",IIf([Support]>(-8.14*Log([Stability])+39.991,"2",IIf([Support]>(-5.061*Log([Stability])+20.856,"3"))))))

Thank you

[Rod]

IIf([Support]>(-15.55*Log([Stability])+75.55,"1",IIf([Support]>(-8.14*Log([Stability])+39.991,"2",IIf([Support]>(-5.061*Log([Stability])+20.856,"3"))))))

According to Help the Iif function takes three arguments - none of which is optional. You have a nested Iif function. Certainly the last Iif has only two arguments (red highlight). I suggest you examine the remainder of the expression.

[June7]

Appears to be unnecessary parens. Try removing the parens highlighted in red. Also need value if last expression false. Why numbers in quote marks?

Probability: IIf([Support]>(-15.55*Log([Stability])+75.55,"1",IIf([Support]>(-8.14*Log([Stability])+39.991,"2",IIf([Support]>(-5.061*Log([Stability])+20.856,"3",0))))))

Provide some sample data for Stability.

[twisted]

Appears to be unnecessary parens. Try removing the parens highlighted in red. Also need value if last expression false. Why numbers in quote marks?

Probability: IIf([Support]>(-15.55*Log([Stability])+75.55,"1",IIf([Support]>(-8.14*Log([Stability])+39.991,"2",IIf([Support]>(-5.061*Log([Stability])+20.856,"3",0))))))

Provide some sample data for Stability.

Thank you June7 you pointed out my error. I have a working formula now.