Water_foul wrote: > i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is > there a icq chatroom for php?) > "Shahmat Bin Dahlan" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] >> 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); >> >
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/) To unsubscribe, visit: http://www.php.net/unsub.php
