Re: Another newbie question
| From: | Pete Holsberg | Date: | Mon, 06 Sep 2004 22:32:35 +0000 |
| Subject: | Re: Another newbie question | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-36078@lists.php.net to get a copy of this message | ||
> --- Pete Holsberg <pjh@mccc.edu> wrote:
>
> > On Mon, 6 Sep 2004
> > php-list-replies@listmail.innovate.net wrote:
> >
> > > you should copy/paste that into a command-line
> > msyql
> > > session and see if that works.
> >
> >
> > I cut the query back to
> >
> > SELECT * FROM Directory WHERE
LastName LIKE Joe
> > ORDER BY Street, HouseNum, LastName
> >
> > and ran it manually. I got
> >
> > ---
> > SQL-query :
> >
> > SELECT *
> > FROM Directory
> > WHERE LastName
> > LIKE Joe
> > ORDER BY Street, HouseNum, LastName
> > LIMIT 0 , 30
> >
> > MySQL said:
> >
> > Unknown column 'Joe' in 'where clause'
> > --------
> >
> > Perhaps I don't understand LIKE. LastName is a
> > column; Joe
> > is a value. ???
On Mon, 6 Sep 2004, Stuart Felenstein wrote:
> I think you need to put '' around joe 'joe'
>
> Stuart
Tried that. It said that 'Joe' was an unknown column!
Besides, Joe is the value of a variable passed by another
script.
Thanks.
> >
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> >
> >
>
>