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Supplied argument is not valid MySQL result resource?

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  #1  
Old July 24th, 2003, 07:54 AM
Gee Gee is offline
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Supplied argument is not valid MySQL result resource?
Hi there everyone,

Long time ive been away! now im back on the php trail doing a little revision.

Anyway, I have created the following username / password login script:

PHP Code:
<?
extract($_Post);
$link = mysql_connect("localhost", "uname", "pass");
mysql_select_db("auth", $link);

$name = mysql_query("SELECT name FROM logstudent WHERE name = '$logon' AND pass = '$password'", $link);
$num_rows = mysql_num_rows($name);

if ($num_rows <= 0)
        {
             echo "<p align=center>You either entered your user name or password wrong. </p>";
exit;
        }
echo "<p align=left>You are now logged in</p>";
include ("showdetails.php");
?>


I keep getting the following warning: -

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\www\checkstudent.php on line 8
(line 8 is : $num_rows = mysql_num_rows($name);

I know that I am supplying the correct username and password, so it has to be something with the actual database/table?

Can anyone explain what this error means so I can track down the problem further.

Also, the database is called "auth", and there are 2 columns called: - "logon" for usernames ans "password" for passwords"

As always, thanX in advance.
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Last edited by Gee : July 24th, 2003 at 08:11 AM.
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  #2  
Old July 24th, 2003, 08:57 AM
crazytrain81 crazytrain81 is offline
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Don't extract your superglobals, it makes for unreadable and unmanageable code in the long term.

If you're having problems with a query, tack on what i've tacked onto the mysql_query() below to see the error AND your sql statement so you can compare the error message to what you're trying to send the db.
PHP Code:
<?
$link = mysql_connect("localhost", "uname", "pass");
mysql_select_db("auth", $link);

$sql = "SELECT name FROM logstudent WHERE name = '".$_POST['logon']."' AND pass = '".$_POST['password']."';";

$name = mysql_query($sql, $link) or die(mysql_error().'<br> SQL:'.$sql);
$num_rows = mysql_num_rows($name);

if ($num_rows <= 0)
        {
             echo "<p align=center>You either entered your user name or password wrong. </p>";
exit;
        }
echo "<p align=left>You are now logged in</p>";
include ("showdetails.php");
?>
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  #3  
Old July 24th, 2003, 10:01 AM
Gee Gee is offline
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Thanks crazytrain, that got to the bottom of my problem right away! I will use that method from now on.
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