Re: problem....
| From: | water_foul | Date: | Sat, 26 Jun 2004 15:21:52 +0000 |
| Subject: | Re: problem.... | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-34984@lists.php.net to get a copy of this message | ||
sorry i didn't respond quicker but i found out i hadn't uploaded my db to
the server :( sorry for wasting your time
"Richard Hutchins" <Richard.Hutchins@Getingeusa.com> wrote in message
news:1EA7D3AE70ACD511BE6D006097A78C1E05887A05@USROCEXC...
> Haven't seen that this has been solved yet so I thought I'd throw my two
> cents into the fray. First of all, I'd second the earlier suggestion that
> you assign your query to a variable. And, second, I'd recommend that you
> echo out your query to the browser just before it is submitted to the
> database server. That way, you can actually _see_ what's being sent to the
> server. I have found this to be an invaluable troubleshooting tactic and
it
> may help you find the root cause of your invalid MySQL result.
>
>
> > -----Original Message-----
> > From: David Robley [mailto:robleyd@ozemail.com.au]
> > Sent: Wednesday, June 23, 2004 2:53 AM
> > To: php-db@lists.php.net
> > Subject: Re: [PHP-DB] problem....
> >
> >
> > Water_foul wrote:
> >
> > > 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
> > >> > > >
> > >> >
> >
> > news:16692.65.203.232.102.1087950853.squirrel@webmail.pacifier.com....À²²4¯¢Ô
> > -d¡3ׯ>
> > >> >> 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);
> > >> >
> >
> > Two suggestions for debugging mysql problems:
> >
> > 1) Create your query as a variable so you can echo it and see
> > exactly what
> > is being passed to mysql
> >
> > $query = "SELECT Rune, username FROM RuneRunner, RuneRunner_1
> > WHERE User_ID
> > = 3";
> >
> > 2) Use mysql_error() to get the actual error from mysql
> > $result = mysql_query($query,$connection);
> > echo mysql_error();
> > $pic=mysql_fetch_array($result);
> >
> > --
> > David Robley
> >
> > "I'd like to learn a new card game," Tom said wistfully.
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
> >