Re: PROBLEM WITH LIKE OPERATOR IN SELECT WHERE CLAUSE

Already did, however since the field has an index on it, it causes oracle not to use the index and thus it hinders performance. Trying to force oracle to use the index by providing it with a hint, didn't help. thanks very much for answering :)

"Randall Roberts" <randall_at_filer.org.nospam> wrote in message news:3b0dbc3b_3_at_news.pcmagic.net...
> Try
>
> SELECT * FROM table WHERE RTRIM(field, ' ') LIKE 'A%C'
>
> Best!
>
> Randall
>
>
> shamai markel <shamaim_at_amdocs.com> wrote in message
> news:990615107.590810_at_sointnews.amdocs.com...
> > Hi,
> > 1. In difference from '=' operator, LIKE opearator does not ignore
 trailing
> > blanks in CHAR type fields.
> > 2. This means that for a CHAR(5) field containing 'ABC ', the following
> > SELECT statement will return no lines :
> > SELECT * FROM table WHERE field LIKE 'A%C', whereas following SELECT
> > statement will return the line :
> > SELECT * FROM table WHERE field = 'ABC'.
> > 3. Does anyone have a bypass to this problem that doesn't require to pad
> > blanks to the searched string and that doesn't inluence performance ?
> > thanks very much
> > shamai
> >
> >
>
>
Received on Wed Jun 06 2001 - 10:25:47 CEST