Re: problem....
| From: | Cole S. Ashcraft | Date: | Wed, 23 Jun 2004 01:48:46 +0000 |
| Subject: | Re: problem.... | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-34877@lists.php.net to get a copy of this message | ||
The error he is getting means that his MySQL resource (in this case the
results of the query) does not exist. Try
$pic1=mysql_fetch_array($pic);
Also, can you give the exact text of the error (line number and all) and line numbers for the entire passage?
Cole
Shahmat Bin Dahlan wrote:
Your SQL statement:RuneRunner> >> > RuneRunner_1 WHERE (User_ID = 3)',$connection);'SELECT Rune, username FROM RuneRunner RuneRunner_1 WHERE (User_ID = 3)'What is "RuneRunner" and "RuneRunner_1"? Are these two different tables. If it is, you might want to separate them with a comma. Why not try getting rid of the brackets surrounding "User_ID=3"? And also wrap single quotes around the digit "3". ----- Original Message ----- From: "water_foul" <phpdbnews@aaron.aichlmayr.net> Date: Wednesday, June 23, 2004 8:58 am Subject: Re: [PHP-DB] problem....I checked em all they were right "Daniel Clark" <dclark@nwlink.com> wrote in message news:18323.65.203.232.102.1087951969.squirrel@webmail.pacifier.com...Sounds like it doesn't like your SQL statement. Perhaps a fieldor tablename is incorrect?MySQL> > resultWarning: mysql_fetch_array(): supplied argument is not a validnews:16692.65.203.232.102.1087950853.squirrel@webmail.pacifier.com...>resource in ....... "Daniel Clark" <dclark@nwlink.com> wrote in messageWhat error are you getting?why doesn't this work:$pic=mysql_query('SELECT Rune, username FROM
-- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.$pic=mysql_fetch_array($pic);