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Message Posted: Mon, 03 Sep 2001 @ 16:32:04 GMT


     
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Subj:   Re: Like comparison in macro
 
From:   Ulrich Arndt

Hi,

I suggest that the result is correct. You defined the input parameter description as char(80). I suggest that even if your input is only '%' the like comparison will be done by '% ' (79 blanks). Try to define the description also as varchar and use the trim function description variables trim(description) like trim(:description) ).


Regards

Ulrich



     
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