将http响应代码从PHP返回到AJAX

Question

我想做一个网站的登录页面。我有一个函数，它使用AJAX向PHP脚本发送请求，以检查是否输入了正确的用户名和密码。如果查询返回成功的结果，则发送http_response_code（200），否则发送http_response_code（403）。然而，登录功能似乎没有返回任何响应状态。反应似乎是不确定的。在这种情况下，即使输入了正确的密码和用户名，该函数也会给我错误密码或用户名的窗口警报。我应该检查什么条件来确定基于超文本传输协议响应代码的成功函数应该做什么？是否有其他方法可以根据PHP脚本的功能将条件返回到AJAX？

这里是登录函数的代码。

function login(){
var username = document.getElementById("username").value;
var password = document.getElementById("password").value;
var dataString = 'username1=' + username + '&password1=' + password;
if (username == '' || login == ''){
    window.alert("Please fill in username or password.");
}
else{
    $.ajax({
        type: "POST",
        url: "login.php",
        data: dataString,
        cache: false,
        crossDomain : true,
        success: function(response) {
            if (response.status == 200) {
                window.location = 'http://localhost/site.html';
            }
            else {
                window.alert("The password or username you have entered is not valid");
            }
        }
    });
}        
return false;

}

这是我的php脚本。

<?php
header('Access-Control-Allow-Origin: *');
header('Access-Control-Allow-Methods: GET, POST, PATCH, PUT, DELETE, OPTIONS');
header('Access-Control-Allow-Headers: Origin, Content-Type, X-Auth-Token');
$password2 = $_POST['password1'];
$username2 = $_POST['username1'];
$connection = mysqli_connect("localhost", "root", "password", "database") or die("Unable to connect to MySQL");  
$query = mysqli_query($connection, "SELECT * FROM users where username = '$username2' AND password = '$password2'") or die(mysqli_error($connection));
$row = mysqli_fetch_array($query, MYSQLI_BOTH) or die(mysqli_error($connection));
if(!empty($row['username']) AND !empty($row['password'])) {
    session_start();
    $_SESSION['username'] = $username2;
    http_response_code(200);
    echo "Successful Login";
    exit;
}
else{
    http_response_code(403);
    echo "The password or username you have entered is not valid";
}
mysqli_close($connection);
?>

Answer 1

在成功函数中使用的响应参数是AJAX调用返回的数据，而不是状态和/或标头。您可以通过获取函数的第二个参数来获得描述状态的字符串：

$.ajax({
  ...
  success: function(data, status){
    if(status == "success"){
      // success code.
    }
  }
});

查看jQuery。有关更多信息，请参阅ajax文档。

Answer 2

当您检查响应并发送get响应时。状态您手中实际上没有数组或对象作为响应：

因此，在检查登录时，您可以创建一个带有状态和消息和json_encode（）的数组，这样javascript代码就可以提取并读取它。

<?php
// fix your query connection - you are currently vulnerable. It does go outside of the scope of your question so I am not going to tackle it here.
if(!empty($row['username']) AND !empty($row['password'])) {
    session_start();
    $_SESSION['username'] = $username2;
    $return = array(
        'status' => 200,
        'message' => "Login Successful."
    );
    http_response_code(200);
}
else{
    $return = array(
        'status' => 403,
        'message' => "Login attempt denied."
    );
    http_response_code(403);
}
print_r(json_encode($return));

现在，您可以在AJAX函数中获得响应：

    success: function(response) {
        var data = $.parseJSON(response);
        if (data.status == 200) {
            window.location = 'http://localhost/site.html';
        }
        else {
            window.alert(data.message);
        }
    }