Re: What is wrong with next code

[John Holmes <holmes072000@xxxxxxxxxxx>]

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.

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---John Holmes...

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