lorsque j’exécute ce code il affiche ce message :

error ;Warning: mysql_query() expects parameter 2 to be resource,
Code : Sélectionner tout - Visualiser dans une fenêtre à part
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
<?php
function db_query($db_name, $sql) {
	$sql = str_replace("# ", "", $sql); // basic shield against sql injections
	$sql = str_replace("#' ", "", $sql);
	global $db_connection_type, $db_server_address, $db_user, $db_password;
	switch($db_connection_type) {
		case "odbc":
			$db_connection = odbc_connect($db_name, $db_user, $db_password);
			$result = odbc_exec($db_connection, $sql);
			break;
		case "mysql":
			$db_connection =  new mysqli($db_server_address, $db_user, $db_password);
			$result = mysql_query ($sql, $db_connection) or die (mysql_error()); ;
			// ($db_name, $sql, $db_connection);
	}
	return $result;
}
function fetch_array($array) {
	global $db_connection_type;
	switch($db_connection_type) {
		case "odbc":
			$result = odbc_fetch_array($array);
			break;
		case "mysql":
			$result = mysql_fetch_array($array);
	}
	return $result;
}
?>
Merci