Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resourc

[joshgarrod]

Hi, I am having an issue with a piece of code. I am getting the following error, any ideas please?

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ...

 

<?php
			include "scripts/connect.php";

			$last = $_GET['ref']; // This will be the ID of the page

			echo "the ref number is $last";

			$query = mysql_query("SELECT * FROM news WHERE `ref` = '$last'");

			if(mysql_num_rows($query)==0){

				die("Something went wrong, contact Administrator!");

				  }else{

					while($info = mysql_fetch_array($query)){
					$title = $info ['title'];

					echo "<p><a href=\"..news.php\" target=\"_blank\">$title >></a></p>";

					}
					}
		?>

[cpd]

$query = mysql_query("SELECT * FROM news WHERE `ref` = '$last'");

 

This is where your error is. Ensure everything is spelled correctly etc.

[soma56]

Change:

 

echo "the ref number is $last";

 

to

 

echo "the ref number is ".$last;

 

This isn't what's causing the error though.

 

And as far as the mysql statement:

 

$query = mysql_query("SELECT * FROM news WHERE `ref` = '$last'");

 

I'd take out the ` and either remove them completely or change them to '

 

$query = mysql_query("SELECT * FROM news WHERE ref = '$last'");

 

Make sure the table name (news) and the column within the table (ref) match perfectly in the DB. 

[PFMaBiSmAd]

That error means that your query failed due to an error of some kind (no database connection, permission problem, wrong table/column names, un-escaped data that breaks the sql syntax/allows sql injection,...) If you use mysql_error(), php/mysql will tell you why it failed. You can temporarily change your mysql_query() statement to the following to get some debugging information from it -

 

$query = mysql_query("SELECT * FROM news WHERE `ref` = '$last'") or die('Query failed: ' . mysql_error());

[AyKay47]

Change:

 

echo "the ref number is $last";

 

to

 

echo "the ref number is ".$last;   

 

this is not needed, variables are interpolated inside of double quotes, what the OP has is fine.

 

I'd take out the ` and either remove them completely or change them to '

 

why? back-ticks are perfectly valid and are typically encouraged to wrap identifiers.

 

OP, debug the query a little to see what is going wrong:

 

$sql = "SELECT * FROM news WHERE `ref` = '$last'";
$query = mysql_query($sql);
if(!$query)
{
    echo $sql . "<br />" . mysql_error();
}

[joshgarrod]

Thanks for your help everyone, problem solved. Stupid typo on DB name.