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Topic: appel aux connaisseurs mysql et php
Posted by: Bussiere at jeu. 19 sept. 2002 18:20:02 CEST

Keywords: pfffffffffff ni peng

Message:

comment je fais pour que mon serveur sql il accepte d'autre user ?

dans mon .ini du sql g :
[WinMySQLadmin]
Server=E:/mysql/bin/mysqld-nt.exe
user=Bussiere
password=jdrrp5


mon code php est :
<?
echo "<html>";
echo "</html>";
mysql_connect('iut2a.dns2go.com', 'Bussiere', 'jdrrp5'); // 1
mysql_create_db('harem');
mysql_select_db('harem',$db);
mysql_query('CREATE TABLE user (log VARCHAR (15) not null,pass VARCHAR (15) not null)');
mysql_close($db); // 6

?>
et mon resultat est :
Warning: Access denied for user: 'Bussiere@melody' (Using password: YES) in E:\site\harem\base.php on line 4

Warning: MySQL Connection Failed: Access denied for user: 'Bussiere@melody' (Using password: YES) in E:\site\harem\base.php on line 4

Warning: Supplied argument is not a valid MySQL-Link resource in E:\site\harem\base.php on line 6

Warning: Supplied argument is not a valid MySQL-Link resource in E:\site\harem\base.php on line 8
bref je nage .....

merci
@+
Bussiere




Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/clients/13eaf4559a54d78787520f07cab62616/web/panda/archreply.php on line 98

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/clients/13eaf4559a54d78787520f07cab62616/web/panda/archreply.php on line 101
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