Ok, I aplogize, for I neglected to notice that you were not passing back the db connection $link variable from your openDB function, so here is what you are going to want to do.
page connectie.php
Function openDB()
{
$link = mysql_connect("localhost", "host", "pas") or die("geen connectie mogelijk : " . mysql_error());
print "connectie ok";
mysql_select_db("database") or die("geen database gevonden");
return $link;
}Function closeDB($ref_link)
{
mysql_close($ref_link);
}include ('connectie.php');
//open connectie
$connectie = openDB();
//code
//close connectie
closeDB($connectie);
Jordan S. Jones
alain dhaene wrote:
I have try but I get the following error:
Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in /home/schoolre/public_html/Hitek/Online/connectie.php on line 16
Alain
"Jordan S. Jones" <[EMAIL PROTECTED]> schreef in bericht
news:[EMAIL PROTECTED]
Alain,
You can pass it in the function as a variable.
E.g...
Function closeDB($ref_link) { mysql_close($ref_link); }
closeDB($connectie);
Hope this Helps,
Jordan S. Jones
--
I am nothing but a poor boy. Please Donate..
https://www.paypal.com/xclick/business=list%40racistnames.com&item_name=Jord an+S.+Jones&no_note=1&tax=0¤cy_code=USD
-- I am nothing but a poor boy. Please Donate.. https://www.paypal.com/xclick/business=list%40racistnames.com&item_name=Jordan+S.+Jones&no_note=1&tax=0¤cy_code=USD
-- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
