PHP mysql_connect不返回布尔值

Question

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

Answer 1

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_connect since 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()

Answer 2

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);
?> 

Answer 3

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) .