Posted by Schraalhans Keukenmeester on 04/27/07 13:02

On Fri, 27 Apr 2007 04:53:04 -0700, Kurda Yon wrote:

> It is me again. It seems that I solve the previous problem (with your
> help) but get a new one (which seems to be very related to the
> previous one).
>
> This part of code produce a message about mistake:
>
> $link = mysql_connect( "localhost","tmp_user","pswxxx" );
> function fff()
> {
> $link_new = mysql_connect( "localhost","tmp_user","pswxxx", true );
> mysql_close( $link_new );
> }
> fff();
> mysql_select_db( "sss", $link );
> $result = mysql_query( "select * from root" );
>
> The message is:
> Access denied for user: 'nobody@localhost' (Using password: NO)
> A link to the server could not be established
>
> It complains about the last line. And this message disappears if I do
> not coll the function fff (I replace fff(); by //fff();).
>
> Can anybody explain this behavior? It seems that connection made to
> the database in the function somehow influences on other connection.

I'm guessing here: You use literal strings here in your post making the
mysql connection, but do you perhaps use variables in your real code? If
so, (and assuming globals are off) the function doesn't see these
variables unless you specify them global in the function.

Other than that, I can't think of anything right now. I cannot reproduce
the error anyway with my code (below).

<?PHP
require 'dbdata.inc.php';
$link = mysql_connect($dbhost,$dbuser,$dbpasswd);
function dbopen() {
global $dbhost, $dbuser, $dbpasswd;
$new_link = mysql_connect($dbhost,$dbuser,$dbpasswd,true);
mysql_close($new_link);
}
dbopen();
mysql_select_db($dbname,$link);
$result= mysql_query('select count(*) from customers',$link);
$row = mysql_fetch_row($result);
echo $row['0']; // outputs 13, oughtta be closer to 50 if you ask me;-)
?>

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