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I have the following PHP code:
$con = mysql_connect("localhost","name","pass") or die(mysql_error());
$db = "db";
mysql_select_db($db,$con);
Now in my experience, $con should be true or false. When I echo $con I get:
Resource id #25
If I do the following code, the echo never fires (as to be expected after the above statement):
if($con) { echo "it worked"; }
When I run a query against this connection, everything works as expected. Is there a reason why this $con will not be true or false?
What am I doing wrong?
Thanks

Check mysql_connect Return Values :
Returns a MySQL link identifier on success or FALSE on failure.
So to check the connection:
if($con !== false) { echo "it worked"; }
or to quit in case of an error:
if (!$con) {
die('Could not connect: ' . mysql_error());
}
A word of advice though, better to use the MySQLi or PDO_MySQL instead of mysql_connectsince it will soon be deprecated:
Warning
This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include:
mysqli_connect()
PDO::__construct()

Its simply b/c of the or or die(mysql_error()); at the end.
Give this a shot:
<?php
$con = mysql_connect("localhost","name","pass");
$db = "db";
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db,$con);
?>

I think it is because of the or die(mysql_error()); part. I should be the $con = mysql_connect("localhost","name","pass") only. Then you can check like this: if (!$con).