Re: Like in embedded SQL

[Vern Hamberg <vhamberg@xxxxxxxxxxx>]

If you column were varying length and filled with a trimmed value, you 
would not need the trailing wildcard, because it would have stopped 
processing at the length of the value. But you can't count on that, 
right? So either the strip() function or the trailing wildcard will do 
the trick - of course, this means that a table scan is likely to take place.

Vern

Joep Beckeringh wrote:
> Terri,
>
> Assuming LLOC is 6 characters long, it contains '2A200 '. The trailing 
> space is your problem; use TRIM(LLOC) and you should be fine.
>
> Joep Beckeringh
>
>
> Op 29-12-09 20:39, THarteau@xxxxxxxxxxxxxxxxxx schreef:
>   
> > Hi,
> >          Are there any issues when using LIKE in an SQL declare statement?
> > I have the following line:
> > /EXEC SQL
> > + DECLARE C6 CURSOR FOR SELECT LPROD, LWHS, LLOT, LLOC, LOPB, LISSU,
> > +  LADJU, LRCT, LIALOC, LISTAT, WLSTS FROM ILIL01 JOIN ILM ON WWHS =
> > +  LWHS AND WLOC = LLOC WHERE LWHS = :KEYWHS AND LPROD = :PROD AND L
> > + LOC LIKE '%200' ORDER BY WLSTS, LLOC FOR FETCH ONLY
> > /END-EXEC
> >
> > It seems to work when it finds 11A200 or 22A200, but it is not returning
> > 2A200. Any ideas? Thanks!
> >
> > <===================================================>
> >
> > Terri Harteau
> > Felker Brothers Corporation
> > ****************
> > "Do not follow where the path may lead. Go instead where there is no path
> > and leave a trail."
> > Ralph Waldo Emerson
> >
> > ****************
> >
> >
> >
> >    
> >     
>
>