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Embedding Left( ) into FindFirst criteria...

931 Recognized Expert Contributor
This is not a question, but rather an informational post about something I just solved. My issue was that I'm trying to use

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  1. rst.FindFirst (str1 & " = '" & txtLastNameSearch.Value & "'")
My goal was to search through a table with a field fldNameLast, but only compare the leftmost int1 characters in this field to my text box value (txtLastNameSear ch.Value). So originally I tried something like

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  1.  str1 = "Left([fldNameLast],int1)" 
But the VB compiler complained about the use of int1 (which is set earlier in the code). After playing around with it for a while, I settled on

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  1.  str1 = "Left([fldNameLast]," & int1 & ")"
And this works very well, at least for all the values of txtLastNameSear ch.Value that I've tried so far. Seems like the main rule is to keep variables outside the quotes...
Nov 14 '07 #1
1 1550
12,516 Recognized Expert Moderator MVP
Thanks for sharing. What you ran into was a variable scope issue. Not all forms and functions can see every other forms' and functions' variables, which can be confusing to the uninitiated. Hopefully this will help other potential readers with problems.
Nov 14 '07 #2

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