i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is there a icq chatroom for php?) "Shahmat Bin Dahlan" <shahmatd@xxxxxxxxxxxx> wrote in message news:16780f91a.f91a16780@xxxxxxxxxxxxxxxxxxxxxxxxx > 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@xxxxxxxxxxxxxxxxxxx> > Date: Wednesday, June 23, 2004 8:58 am > Subject: Re: problem.... > > > I checked em all they were right > > "Daniel Clark" <dclark@xxxxxxxxxx> wrote in message > > news:18323.65.203.232.102.1087951969.squirrel@xxxxxxxxxxxxxxxxxxxxxxx > > > 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@xxxxxxxxxx> wrote in message > > > > > > news:16692.65.203.232.102.1087950853.squirrel@xxxxxxxxxxxxxxxxxxxxxxx> > >> 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 > > > > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
