PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Question

After uploading the site to my web server I get this message:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource 
in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 8

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource 
in /home1/m1k3ey/public_html/MikeyDev.com/teamdesire/playersheet.php on line 9

My PHP version is:

PHP Version 5.2.17

I cant see where im going wrong in my code, can anyone please assist:

          mysql_select_db('teamdesire', $link);
          $query = "SELECT * FROM playershowercase";
          $result = mysql_query($query,$link);
          $row = array();
Line 8 >  while($row[] = mysql_fetch_array($result));
Line 9 >  $count = mysql_num_rows($result);
          $random = rand(0,$count-1);

As per the usual for your error message: Your query has failed somehow, your code blindly assumes success, and blundered onwards. — Marc B, Jul 16 '13 at 15:22
@MikeyT - have you tried `$result = mysql_query($query,$link) or die (mysql_error());` ? Or `var_dump($result`? — andrewsi, Jul 16 '13 at 15:27
Yes ive tried that, in my error log it says exactly the same error which is displayed on screen — MikeyT, Jul 16 '13 at 15:29

Malcolm · Answer 1 · 2013-07-16T16:25:07.740

Maybe something like the following?

      mysql_select_db('teamdesire', $link);
      $query = "SELECT * FROM playershowercase";
      $result = mysql_query($query,$link);
      $row = mysql_fetch_assoc($result);
      $count = mysql_num_rows($result);
      $random = rand(0,$count-1);

Or

      mysql_select_db('teamdesire', $link);
      $query = "SELECT * FROM playershowercase";
      $result = mysql_query($query,$link);
      while ($row = mysql_fetch_array($result, MYSQL_NUM))
      $count = mysql_num_rows($result);
      $random = rand(0,$count-1);

score 0 · Answer 2 · answered Jul 16 '13 at 16:49

Possible error sources:

Db is not running on specified socket or port.
Insufficient permissions on the table or database.
Incorrect login credentials.
Table does not exist or is misspelled.

Echo line per line when you troubleshoot.

is_resource($link) or die('Could not connect');
mysql_query(...);
echo mysql_error();

score -1 · Answer 3 · answered Jul 16 '13 at 15:30

-1

Try this:

          mysql_select_db('teamdesire', $link);
          $query = "SELECT * FROM playershowercase";
          $result = mysql_query($query,$link);
          $row = array();
          while($result = mysql_fetch_array($result))
          {
          $row[]=$result;
          }
          $count = mysql_num_rows($result);
          $random = rand(0,$count-1);

answered Jul 16 '13 at 15:30

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I get the same error again unfortunately – MikeyT Jul 16 '13 at 15:36

PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

3 Answers3