Warning: mysql_num_rows()

I am getting this error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/myuser/public_html/saml/oc.php on line 35

Code (markup):

Line 35 is shown below

$username=$_SESSION['username'];
$above = mysql_query("SELECT * FROM users WHERE username='$username'");
$fetch = mysql_fetch_object($above);


$oc_id=strip_tags($_GET['oc_id']);
if (strip_tags($_GET['place']) == "we"){
$use="we_inv";
$a="we";
$query= "SELECT * FROM oc WHERE we_inv='$username' AND id='$oc_id'";
}elseif (strip_tags($_GET['place']) == "ee"){
$use="ee_inv";
$a="ee";
$query= "SELECT * FROM oc WHERE ee_inv='$username' AND id='$oc_id'";

}elseif (strip_tags($_GET['place']) == "driver"){
$use="driver_inv";
$a="driver";
$query= "SELECT * FROM oc WHERE driver_inv='$username' AND id='$oc_id'";
}

$round=mysql_query($query);
$check=mysql_num_rows($round); <----LINE 35 HERE
if ($check != "0"){

mysql_query("UPDATE `oc` SET `$a`='$username' WHERE `id`='$oc_id'");
mysql_query("UPDATE `users` SET `oc`='1' WHERE `username`='$username'");
echo "<SCRIPT LANGUAGE='JavaScript'>
window.location='oc.php';

</script>";
}

PHP:

 

There may be an error in the query thats why $round isn't a valid resource try editing line 34 to:

$round=mysql_query($query) or die(mysql_error());

PHP:

to see if there is an error.

Cheers!

 

Yes thanks.

I got the following error

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #4' at line 1

Code (markup):

 

It's likely caused by one of your user submitted variables not being set/defined, theirfore I've added a few if statements to check - also improved your code slightly (to make it more readable)...use the following code, and let me know how that goes (also quote any errors - if any; within your next post):

<?php
error_reporting(E_ALL);
if (!isset($_SESSON['username'])) {
    echo 'You must be logged in!';
} elseif (!isset($_GET['oc_id'])) {
    echo 'oc_id is not set!';
} elseif (!isset($_GET['place'])) {
    echo 'place is not set!';
} else {
    $username = $_SESSION['username'];
    $above    = mysql_query("SELECT * FROM users WHERE username = '{$username}'");
    $fetch    = mysql_fetch_object($above);

    $oc_id = intval($_GET['oc_id']);
    if ($_GET['place'] == 'we') {
        $use   = 'we_inv';
        $a     = 'we';
        $query = "SELECT * FROM oc WHERE we_inv = '{$username}' AND id = '{$oc_id}'";
    } elseif ($_GET['place'] == 'ee') {
        $use   = 'ee_inv';
        $a     = 'ee';
        $query = "SELECT * FROM oc WHERE ee_inv = '{$username}' AND id = '{$oc_id}'";
        
    } elseif ($_GET['place'] == 'driver') {
        $use   = 'driver_inv';
        $a     = 'driver';
        $query = "SELECT * FROM oc WHERE driver_inv = '{$username}' AND id = '{$oc_id}'";
    }
    
    $round = mysql_query($query);
    $check = mysql_num_rows($round); 
    if ($check != 0):
        mysql_query("UPDATE oc SET {$a} = '{$username}' WHERE id = '{$oc_id}'");
        mysql_query("UPDATE users SET oc = '1' WHERE username = '{$username}'");
?>
<script language='JavaScript'>
window.location='oc.php';
</script>
<?php
    endif;
}
?>

PHP:

 

That didn't work but I noticed that I used $query twice, so i changed it to $getit and now get the error below

Query was empty

Code (markup):

 

Erm that isn't an error? (or atleast from my entire time doing PHP havn't ever come accross that :/)

Your current code sets a variable $query based on the $_GET...and then later on in the script you use that $query.

So if those $_GET's are not set you'll get an error as their is no query to execute.

The same applies for the $_SESSION['username']

Theirfore:

- make sure all variables exist/are set so it can query properly.

- make sure all column names and table names are exist!