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:[EMAIL PROTECTED] > Sent: Wednesday, June 23, 2004 2:53 AM > To: [EMAIL PROTECTED] > 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" <[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 > > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
