Re: problem....
| From: | Shahmat Bin Dahlan | Date: | Wed, 23 Jun 2004 01:27:28 +0000 |
| Subject: | Re: problem.... | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-34875@lists.php.net to get a copy of this message | ||
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...•˜¤v
> ^‹V%žz Ç‹
> > 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
> > >
>
> news: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);
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>