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";
> }
> -------------------------------------------------
> [EMAIL PROTECTED]:~/src/PHP# php -q mysql_connect3.php
> Result: 2
>
>
>
> On Thu, 28 Oct 2004 23:08:39 -0400
> John Holmes <[EMAIL PROTECTED]> 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
> >
> >
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