The IIf() function should contain three arguments:
IIf( [Condition to Evaluate], [Return if True], [Return if False])
I will sometimes set the third argument to Null.
Duane Hookom MVP
MS Access
From: MS_Access_Professionals@yahoogroups.com
Why not try IIF(instr(LOB, "FHN") > 0, ....)
Jeff
Hello everyone.
I just have a quick question. I feel like I should know the answer to this but I'm drawing a blank. All that I want to do is use IIF to find a value and if true insert a text phrase. Easy enough. However, the twist. I want to be able to evaluate a field value that contains my text as part of the field value. In case I didn't explain my self correctly, here's an example.
The field [LOB} contains the value "FHN/ACA, ICP only". I want to evaluate this information to see if "FHN" is a part of the value. Then I want Access to return the value "FHN" if it is part of the text. I've tried two different ways to write the function.
IIF([LOB] Like "*FHN*", "FHN")
IIF([LOB] = "*FHN*", "FHN")
They both come back as blank when I run them. Can someone tell me what I'm doing wrong? I seem to be having a "brain fart."
Thanks.
Karen
Posted by: Duane Hookom <duanehookom@hotmail.com>
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