警告：mysqli_real_escape_string（）期望参数1为mysqli，在C：\ xampp \ htdocs \ lr \ lr.php中给出的字符串

Question

I am trying to sanitize all the data of array $register_data . For that I am using array_walk() in which I am passing the second parameter as the sanitize function, but I am getting this warning :

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, string given in C:\xampp..

Below is my code:

function register_user($register_data) {
    array_walk($register_data, 'array_sanitize');
    $register_data['password'] = md5($register_data['password']);
    print_r($register_data);
}

function array_sanitize($conn, &$item) {
    $item = mysqli_real_escape_string($conn, $item);
}

$conn = mysqli_connect($servername, $username, $password, $database);
if(!$conn) {
    die();
}

Answer 1

array_sanitize requires 2 arguments, and the second one should be the item. The arguments array_walk() passes to the function are the array element and its index.

The simplest fix is to change array_sanitize to take just one argument, and get $conn from the global variable:

function array_sanitize(&$item) {
    global $conn;
    $item = mysqli_real_escape_string($conn, $item);
}