Re: What is wrong with next code

Have anyone answers on the "strange behaviour" discussed in the thread ? > > I has noted specially: > It works OK in PHP 4.3.6 or when line "$db2 = 0;" had been commented > in all described versions of PHP. > > So next example works perfect: > ------------------------------------------------- > error_reporting(E_ALL); > $db1 = mysql_connect ('localhost', 'test', '1'); > $db2 = mysql_connect ('localhost', 'test', '1'); > mysql_close($db2); > $result = mysql_query('SELECT 1+1', $db1); > if ($result) > { > $row = mysql_fetch_row($result); > echo "Result: {$row[0]}\n"; > } > ------------------------------------------------- > root@gate:~/src/PHP# php -q mysql_connect3.php > Result: 2 > > > > On Thu, 28 Oct 2004 23:08:39 -0400 > John Holmes <holmes072000@charter.net> wrote: > > > Vladimir Shiray wrote: > > > ------------------------------------------------- > > > Warning: mysql_query(): 4 is not a valid MySQL-Link resource in ... > > > $result = mysql_query('SELECT 1+1', $db1); > > > ------------------------------------------------- > > [snip] > > > $db1 = mysql_connect ('localhost', 'test', '1'); > > > $db2 = mysql_connect ('localhost', 'test', '1'); > > > mysql_close($db2); > > > > If you connect with the same parameters, PHP will just reuse the > > previous connection. So you only have one connection and closing either > > one will result in the connection being lost, hence the "invalid link > > resource" message. > > > > -- > > > > ---John Holmes... > > > > Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/ > > > > php|architect: The Magazine for PHP Professionals _ www.phparch.com > > > >