Subject Re: [firebird-support] 'like' with % at the begin finds nothing
Author Nando Dessena

R> given a small table
R> table t ( v char(10) )
R> with only one entry:
R> insert into t ( v ) values ( 'abcd' )

this is actually 'abcd ', since the column is a char (as opposed
to a varchar).

R> the following select does not find any entry:
R> select * from t where v like '%bcd'

as designed. Strings are padded for comparison (for example when you
use =), but not when using LIKE. This is standard-compliant. You might
want to do one of the following:

- switch from char to varchar
- pad the argument of LIKE with spaces to the column length.

Nando Dessena
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