> -----Original Message-----
> From: Jennifer Downey [mailto:[EMAIL PROTECTED]]
> 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: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
