Question

在用户登录网站时尝试将用户名另存为cookie。

// LOGIN USER
if (isset($_POST['login_user'])) {
  $username = mysqli_real_escape_string($conn, $_POST['username']);
  $password = mysqli_real_escape_string($conn, $_POST['password']);

  if (empty($username)) {
    array_push($errors, "Username is required");
  }
  if (empty($password)) {
    array_push($errors, "Password is required");
  }

  if (count($errors) == 0) {
    $password = md5($password);
    $query = "SELECT * FROM tblmember WHERE Username='$username' AND Password='$password'";
    $results = mysqli_query($conn, $query);
    if (mysqli_num_rows($results) == 1) {
        $cookie_name = "user";
        $cookie_value = $results;
        setcookie($cookie_name, $cookie_value);
      $_SESSION['username'] = $username;
      $_SESSION['success'] = "You are now logged in";
      //set username as a cookie & then use that to access the user pages.
        echo "Cookie '" . $cookie_name . "' is set!<br>";
        echo "Value is: " . $_COOKIE[$cookie_name];
      header('location: profile.php');
    }else {
        array_push($errors, "Wrong username/password combination");
        echo "Cookie named '" . $cookie_name . "' is not set!";
    }
  }
}

我尝试对此进行测试，并且首次使用：$ cookie_value =“ lol”;然后我将其切换为$ cookie_value = $ results;保存更改并从Google chrome中删除该Cookie。然后，我尝试再次在chrome中重新打开该页面，以发现根本没有设置cookie。

Answer 1

setcookie的第二个参数必须是字符串。您正在向它传递mysqli::result对象。您需要从查询中获取结果，然后使用类似serialize的形式将结果转换为字符串，然后可以将其存储到cookie中。例如：

if (mysqli_num_rows($results) == 1) {
    $row = mysqli_fetch_assoc($results);
    $cookie_name = "user";
    $cookie_value = serialize($row);
    setcookie($cookie_name, $cookie_value);
    // ...

PHP：设置Cookie

1 个答案: