RE: [PHP] SQL Warning

> -----Original Message----- > From: Jennifer Downey [mailto:jendowney@attbi.com] > Sent: 24 April 2002 04:46 > > Would you please direct your attention to this URL > > http://testphp.netfirms.com/code1.html > > Look at the bottom where the big orange commented syntax is > and explain what > is going on there? ======================================== $query = "SELECT name FROM {$config["prefix"]}_users WHERE uid={$session["uid"]}"; $ret = mysql_query($query); // this is line 14 while($row = mysql_fetch_array($ret)) . . . // this is the problem area. It is giving me this sql warning Warning: Supplied argument is not a valid MySQL result resource in /home/public_html/sortitems.php on line 14 look at top for line 14 why is it // looking at line 14 I am not asking for any of that information? ======================================== No idea why the error message is so far down the page, but it is certainly referring to line 14 -- it is telling you that the $ret you are supplying to mysql_fetch_array is invalid. Why? Because the mysql_query failed, and so $ret was not assigned valid result resource. This is why you should ALWAYS check the return value of a mysql_query call before trying to fetch its results. Now, why is the query failing? Well, let's take a look at this line: $query = "SELECT name FROM {$config["prefix"]}_users WHERE uid={$session["uid"]}"; The stuff to the right of the "=" consists of: (1) the string "SELECT name FROM {$config[" (2) the constant named prefix (which, I guess, doesn't exist) (3) the string "]}_users WHERE uid={$session[" (4) the constant named uid (ditto!) (5) the string "]}" with NO concatenation operators in between. This line ought to generate multiple warnings and notices -- you must have error_reporting set to suppress all these useful messages, so I suggest you change it to something more useful forthwith (such as E_ALL, or at least E_ALL ^ E_NOTICE). So, the final solution to your problem would be either of the following: $query = "SELECT name FROM {$config[\"prefix\"]}_users WHERE uid={$session[\"uid\"]}"; $query = "SELECT name FROM {$config['prefix']}_users WHERE uid={$session['uid']}"; Hope this helps. Cheers! Mike --------------------------------------------------------------------- Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning & Information Services, JG125, James Graham Building, Leeds Metropolitan University, Beckett Park, LEEDS, LS6 3QS, United Kingdom Email: m.ford@lmu.ac.uk Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211