Error 13 in debug.print

[rmaslia]

I am trying to extract user information from my Active Directory server using ADODB..
Everything works well except for 1 of the attributes which id UID if am using the following statement:

dim strUID
strUID=iff(isnull(adoRecordset.Fields("UID").value),"",adoRecordset.Fields("UID").value)

At Runtime, as soon as there is a non Null value, I receive the following message:
Run-tim error '13': Type mismatch..

I have tried to debug.print but I receive the same error message.

Help would be greatly apprieciated.

Robert

[dglienna]

Add this:

Debug.Print  adoRecordset.Fields("UID").value

Debug.Print  adoRecordset.Fields("UID").value

somewhere and you'll see the problem.

Check the values BEFORE setting the variable

[WoF]

Yes, the problem is the iif() function, I guess.
In an iif always both resulting terms are evaluated. This leads to problems in some case.
Try to formulate this as an If Then Else statement.
Using If Then Else does NOT execute the statements of the opposite condition.

If isnull(adoRecordset.Fields("UID").value) Then strUID = "" Else strUID=adoRecordset.Fields("UID").value

But come to think of it... because of Type Mismatch Error...
If .Fields("UID").Value is a numeric data type you simply have to convert it to string within the Iif statement:

strUID=iff(isnull(adoRecordset.Fields("UID").value),"",CStr(adoRecordset.Fields("UID").value))