Erreur en exécutant une requête mysql_query()

Version imprimable

lorsque j’exécute ce code il affiche ce message :

Citation:

error ;Warning: mysql_query() expects parameter 2 to be resource,

Code:

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<?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

19/03/2015, 17h32
sabotage

Tu fais une connexion mysqli et tu utilises les fonctions mysql ensuite ; ça ne colle pas.

j'ai utilisé le code comme ceci

Code:

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<?php
include('config.php');
function db_query($database_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":
			$db_connection = odbc_connect($database_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) or die("Error " . mysqli_error("error in db_connection"));
			if (mysqli_connect_errno()) {
		printf("Connect failed: %s\n", mysqli_connect_error("pproblem in connect data base"));
		exit();
	}
	$link= mysqli_connect($db_server_address,$db_user,$db_password,$database_name) or die("Error in link");
	$result = mysqli_query($link, $sql); 
	// ($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;
}
?>

mais il affiche ce message d'erreur !:

Citation:

( ! ) Warning: mysqli_connect()

19/03/2015, 22h44
sabotage

A quoi sert la ligne 19 par rapport à la ligne 26 ou inversement ?

oui tu as raison déjà il présente le même principe d'exécution , j'ai oublié de mettre en commentaire

Code:

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function db_query($database_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":
			$db_connection = odbc_connect($database_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) or die("Error " . mysqli_error("error in db_connection"));
 
			if (mysqli_connect_errno()) {
		printf("Connect failed: %s\n", mysqli_connect_error("pproblem in connect data base"));
		exit();
	}
 
			$result = mysqli_query($sql, $db_connection); 
			// ($db_name, $sql, $db_connection);
		}
 
		return $result;
	}

mais il y a toujours un problème au niveau de la fonction mysqli_query() j'en sais pas pourquoi ?

Code:

$result = mysqli_query($sql, $db_connection);

20/03/2015, 08h53
sabotage

Le premier argument c'est l'identifiant de connexion.
20/03/2015, 13h12
syrine01

merci beaucoup monsieur :)