Supplied argument is not a valid MySQL result resource
 

hi,

i write a code like this

$user_id = $hd_cookie['user_id'];
$user_password = $hd_cookie['auth'];


$sql = "SELECT * FROM hd_users WHERE user_id='$user_id' AND user_password='$user_password'";

$r_user = mysql_query($sql);
$num=mysql_num_rows($r_user); //line 24

it shows Warning
Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\helpdeskfinal\includes\auth.php on line 24

when i print the sql statement and run the statement in MySql Window it gives the result.but from here it didn't give the result why?

i am using PHPTRIAD 2.1.1 version is this version supports mysql_num_rows????

please solve my problem

Thanks
Srinivas

Re: Supplied argument is not a valid MySQL result resource
 

It might return an error if the result of the SELECT contained no rows, but I could be wrong. The code looks ok, I'm not sure what the problem could be.

Re: Supplied argument is not a valid MySQL result resource
 

Debugging tip: You mention that you run the statement in MySql window, and it works, but did you actually copy & paste the statement as generated by your code? Right after you build the $sql variable, output it like so:
[php]
echo $sql;
exit();
[/php]
Then copy & paste that output in your MySql query windows and see if it works. You may find your problem.

Re: Supplied argument is not a valid MySQL result resource
 

I had exactly the same problem and I knew for certain I did spell the cellnames and tablename right.

I've found the solution. Simply put all the cellnames and tablename between ``. That's all!

[php]
$sql = "SELECT * FROM `hd_users` WHERE `user_id`='$user_id' AND `user_password`='$user_password'";
[/php]

If this doesn't work, first get the user_id and user_password out of the ``, only hd_users. If that doesn't work either, I don't it, for me it worked :)

grtz, Jero

Re: Supplied argument is not a valid MySQL result resource
 

Thanks very much, I've just find this via google, and its got me out of a tight spot ;)

Thanks again, and sorry to bring up an old topic :(

Re: Supplied argument is not a valid MySQL result resource
 

Thanks, this thread helped me figure out a problem I was having too.

In my case, the problem was a missing semicolon at the end of the SQL statement. Was this also the problem for the original poster, I wonder?

In other words, this:

...snip... AND user_password='$user_password'";

should have been this:

...snip... AND user_password='$user_password';";