javascript - Ajax Post 请求不起作用

Question

您好，我已经创建了一个使用 ajax 方法登录的脚本，我尝试过，但这对我不成功，您可以让我知道为什么我无法使用此方法吗

function login() {
var username = document.getElementById('username').value;
var pass     = document.getElementById('pass').value;

if(username == '' && pass == '') {
    alert('Fields could not be left empty');
} else {
    $.ajax({
        type     : "POST",
        url      : "includes/register.php",
        data     : "command=login&username="+username+"&password="+pass,
        datatype : "json",
        success  :  function(data) {
            if (data.status == 200) {
                alert('Successfull');
                var return_data = data.responseText;
                document.getElementById('messagearea').innerHTML = return_data;
            } else {
                alert('Unsuccessful');
            }
        }
    });
}
}

它应该成功运行警报，但我收到警报('不成功')，这意味着状态不等于 200 谁能帮我解决这个问题

这是我的 php 代码

if(isset($_POST['command']) && $_POST['command'] == 'login') {
    $username = $_POST['username'];
    $pass     = $_POST['pass'];
    $password = md5($pass);

    $check  = mysqli_query($connection, "SELECT * FROM users WHERE username = '$username'");
    $result = mysqli_num_rows($check);
    if($result != 1) {
        echo "<div class='message'>Response Successful</div>";
    } else {
        echo "<div class='message'>Username/Password did not matched</div>";
    }
}

Answer 1

datatype 中的

T 应该大写，

$.ajax({
      type: "POST",
      url: "includes/register.php",
      data: "command=login&username=" + username + "&password=" + pass,
      dataType: "json",

根据服务器代码，您的数据应如下所示

data: "command=login&username=" + username + "&pass=" + pass,

像这样更改成功处理程序中的代码，

document.getElementById('messagearea').innerHTML = data;