Re: problem....
| From: | water_foul | Date: | Wed, 23 Jun 2004 01:44:01 +0000 |
| Subject: | Re: problem.... | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-34876@lists.php.net to get a copy of this message | ||
i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is
there a icq chatroom for php?)
"Shahmat Bin Dahlan" <shahmatd@sains.com.my> wrote in message
news:16780f91a.f91a16780@mail.sarawaknet.gov.my...
> Your SQL statement:
>
> 'SELECT Rune, username FROM RuneRunner
> RuneRunner_1 WHERE (User_ID = 3)'
>
> What is "RuneRunner" and "RuneRunner_1"? Are these two different tables.
> If it is, you might want to separate them with a comma.
>
> Why not try getting rid of the brackets surrounding "User_ID=3"? And
> also wrap single quotes around the digit "3".
>
>
> ----- Original Message -----
> From: "water_foul" <phpdbnews@aaron.aichlmayr.net>
> Date: Wednesday, June 23, 2004 8:58 am
> Subject: Re: [PHP-DB] problem....
>
> > I checked em all they were right
> > "Daniel Clark" <dclark@nwlink.com> wrote in message
> >
> > news:18323.65.203.232.102.1087951969.squirrel@webmail.pacifier.com...
> > > Sounds like it doesn't like your SQL statement. Perhaps a field
> > or table
> > > name is incorrect?
> > >
> > > > Warning: mysql_fetch_array(): supplied argument is not a valid
> > MySQL> > result
> > > > resource in .......
> > > > "Daniel Clark" <dclark@nwlink.com> wrote in message
> > > >
> > H5
> > ×ä ‰´ ÄMÁ¼ûnews:16692.65.203.232.102.1087950853.squirrel@webmail.pacifier.com...>
> >> What error are you getting?
> > > >>
> > > >> > why doesn't this work:
> > > >> > $pic=mysql_query('SELECT Rune, username FROM
> > RuneRunner> >> > RuneRunner_1 WHERE (User_ID = 3)',$connection);
> > > >> > $pic=mysql_fetch_array($pic);
> >
> > --
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> >
> >