What is wrong with next code
| From: | Vladimir Shiray | Date: | Fri, 29 Oct 2004 02:15:48 +0000 |
| Subject: | What is wrong with next code | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-200600@lists.php.net to get a copy of this message | ||
Can one explain me why I has got a result:
-------------------------------------------------
Warning: mysql_query(): 4 is not a valid MySQL-Link resource in ...
$result = mysql_query('SELECT 1+1', $db1);
-------------------------------------------------
from next example.
The example does not work correctly (as I expect) in PHP 4.3.8, 4.3.9, 5.0.2
It works OK in PHP 4.3.6 or when line "$db2 = 0;" had been commented
in all described versions of PHP.
If it's a correct behaviour in new versions of PHP so
how can I free/unset $db2 variable without triggering
such effect?
-------------------------------------------------
error_reporting(E_ALL);
$db1 = mysql_connect ('localhost', 'test', '1');
$db2 = mysql_connect ('localhost', 'test', '1');
mysql_close($db2);
$db2 = 0;
$result = mysql_query('SELECT 1+1', $db1);
if ($result)
{
$row = mysql_fetch_row($result);
echo "Result: {$row[0]}\n";
}
-------------------------------------------------