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" <[EMAIL PROTECTED]> Date: Wednesday, June 23, 2004 8:58 am Subject: Re: [PHP-DB] problem.... > I checked em all they were right > "Daniel Clark" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] > > 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" <[EMAIL PROTECTED]> wrote in message > > > > news:[EMAIL PROTECTED]> >> 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
