Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource

Question

This is the code where I am constantly getting an error. The same include file works on the other pages, I am only getting problem on this page. Here is the error

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /var/www/html/spywgc/adm/ctshell/getproduct/getproduct.php on line 18

And here is the actual code :

<?php require_once('../../../Connections/spyware_adm.php'); ?>
<?php require_once('../../../includes/lib_gen.php'); ?>

<?php
//function for returing status of website
function Visit($url)
{
    echo $url;
    mysql_select_db($database_spyware, $spyware) || die(mysql_error());
    $select_url= "select product_id from sp_url where url like '%{$url}%'";
    echo $select_url;
    $run_url= mysql_query($select_url, $spyware);
    $result_descr = mysql_fetch_assoc($run_url);
    echo $result_descr;
    return $result_descr;

}
?>

possible duplicate of [mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in php](http://stackoverflow.com/questions/1858304/mysql-fetch-assoc-supplied-argument-is-not-a-valid-mysql-result-resource-in-ph) — Lightness Races in Orbit, Sep 16 '11 at 17:50

score 1 · Accepted Answer · answered Apr 15 '11 at 19:23

mysql_select_db() expects the second parameter to be a resource identifier = your connection. The problem is, that you are running this as a function, inside which your connection is not established. You have to start your function with something like this:

function Visit($url)
{
    $spyware = mysql_connect(); // set this to connect properly
    echo $url;
    mysql_select_db($database_spyware, $spyware) || die(mysql_error());
    // the rest of your function goes on ...

score 1 · Answer 2 · answered Apr 15 '11 at 19:40

I would asume that these two vars $spyware and $database_spyware are null because they are defined (if at all) outside of your function's scope and not declared as global. Try to add

global $database_spyware, $spyware

at the beginning of function Visit(..).

score 0 · Answer 3 · answered Apr 15 '11 at 19:22

Did you check if your initial database connection call succeeded? That returns false on failure, as most of the mysql functions do. As well, you haven't checked if the query succeeded. Most likely it failed and returned FALSE, which you then passed into the fetch call:

$database_spyware = mysql_connect(...);
if ($database_spyware === FALSE) {
   die("Connection failed: " . mysql_error());
}

... etc ...

$run_query = mysql_query(...);

if ($run_query === FALSE) {
   die("Query failed: " . mysql_error());
}

should be the bare mininum acceptable error handling.

I am getting that warning even before it executes the query. — vincy, Apr 15 '11 at 19:25

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource

3 Answers3