Resource id #2

stijnbokhove

Hello,

I have this weird output "Resource id #2" when I browse the file with the code that is below. It is strange, because this code DID work on another server. Does anybody know what I'm doing wrong or how I should re-configure my PHP or MySql?

Thanks. Here is the code

<body>
<?
include("db.inc.php");
mysql_connect($server,$username,$password);
@mysql_select_db($database) or die(mysql_error());

$query = "SELECT * FROM solas_rel WHERE $methode LIKE '$keyword%' ORDER BY achternaam ASC";

$result = mysql_query($query);
echo "debug $result";
$num = mysql_numrows($result);
mysql_close();

if ($num == 0) {
echo "Geen resultaten gevonden.";
}

if ($num > 0) {
include("rel_zoek_result.php");
}

Here is the output:
Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in /Users/stijn/Sites/solas/rel_zoek_do.php on line 18

mikesgt2

How about this:

<? php
include("db.inc.php"); 

/* assign the connection to $link */
$link = mysql_connect($server,$username,$password); 

@mysql_select_db($database) or die(mysql_error()); 

$query = "SELECT * FROM solas_rel WHERE $methode LIKE '$keyword%' ORDER BY achternaam ASC"; 
$result = mysql_query($query, $link);  /* link parameter supplied */
$num = mysql_numrows($result); 
mysql_close($link); 

...

?>

Also, you cannot print a result variable. That is where the "Resource id #2" is coming from.

[deleted]

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

means that your query simply was not valid.

most likely the variables inside the query are empty, breaking the syntax. Print the query to see what's going on.

You can print the exact error like this:

<?
$query = "SELECT * FROM solas_rel WHERE $methode LIKE '$keyword%' ORDER BY achternaam ASC";
if (!$result = mysql_query($query))
{
echo 'Query failed:'.$query.'<BR>'.mysql_error();
}
else
{
// query ok, continue processing.
}
?>