WampServer
Apache, PHP, MySQL on Windows
Home forum
Back to WampServer
presentation
Downloads
Addons
Trainings
Alter Way
Go to Topic:
Previous
•
Next
Go to:
Forum List
•
Message List
•
New Topic
•
Search
•
Log In
•
Print View
error using mysql_num_rows()
Posted by:
alanmarcel
(---.user.veloxzone.com.br)
Date: November 03, 2006 03:20PM
I am using the function mysql_num_rows() and I have received the Warning:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\final\resultUsers.php on line 32
Look at my code:
mysql_select_db('item');
$query = "SELECT * FROM item WHERE ".$tipoProcura." LIKE '%".$termoProcura."%'";
$result = mysql_query($query);
$num_results = mysql_num_rows($result); <------------
echo '<p> NĂºmero de itens encontrados: '.$num_results.'</p>';
I think it's a PHP5's problem..
Options:
Reply
•
Quote
Re: error using mysql_num_rows()
Posted by:
CyberSpatium
(71.237.217.---)
Date: November 03, 2006 08:18PM
$conn = mysql_connect ('localhost', 'username', 'password');
mysql_select_db('item', $conn);
$query = "SELECT * FROM item WHERE ".$tipoProcura." LIKE '%".$termoProcura."%'";
$result = mysql_query($query, $conn);
$num_results = mysql_num_rows($result);
CyberSpatium
WAMP English Forum Admin
Options:
Reply
•
Quote
Re: error using mysql_num_rows()
Posted by:
Andre Rotband
(---.PAEMT705.e.brasiltelecom.net.br)
Date: November 03, 2006 09:00PM
Try to print your $query variable.
mysql_select_db('item');
$query = "SELECT * FROM item WHERE ".$tipoProcura." LIKE '%".$termoProcura."%'";
print $query;
$result = mysql_query($query);
And try:
$result = mysql_query($query) or die(mysql_error());
to see if your query have a error.
Post Edited (11-03-06 21:02)
Options:
Reply
•
Quote
Go to:
Forum List
•
Message List
•
Search
•
Log In
Sorry, only registered users may post in this forum.
Click here to login