IIF 2

[kimchavis]

I was hoping someone could help me with this. Im using Access 2000 and I am new to Access. Im doing an iif statement within a query.

Expr6: IIf([expr5]=1,'Registered',IIf([expr5]=3,'Submitted'),iif([expr5] = 6,'LOI Executed')

of course this is coming back an error. it wasnt coming back an error until I added the last statement:
iif([expr5] = 6,'LOI Executed')

What did I do wrong in adding the last part?

Let me know if there is further information needed to solve this.

Thank you!!

 

[mp9]

Try:

Expr6: IIf([expr5]=1,'Registered',IIf([expr5]=3,'Submitted',iif([expr5] = 6,'LOI Executed')))

http://www22.brinkster.com/accessory/

 

[Golom]

And you have dropped the FALSE part of the last IFF

Expr6: IIf([expr5]=1,'Registered',
       IIf([expr5]=3,'Submitted',
       IIF([expr5]=6,'LOI Executed'[COLOR=red],'None of the Above'[/color])))

 

[ZmrAbdulla]

I think you should have the last part of the IIF statement.
ie: If the values are not the matching one then the text should show "SomethingElse"

Expr6: IIf([expr5]=1,'Registered',IIf([expr5]=3,'Submitted',iif([expr5] = 6,'LOI Executed'[b],'SomethingElse'[/b])))

________________________________________________________
Zameer Abdulla
Help to find Missing people
My father was a realistic father; not a vending machine dispense everything I demanded for!!

 

[ZmrAbdulla]

Golom,
Sorry we were typing at the same time. You succeeded ;-)

________________________________________________________
Zameer Abdulla
Help to find Missing people
My father was a realistic father; not a vending machine dispense everything I demanded for!!