Question

我最近一直在和ajax斗争。我正在尝试将用户名提交给“ return.php”。如果成功，我希望它提醒用户名。

但这并不会成功：功能。有人对此有意见吗？会很有帮助的！

    <!DOCTYPE html>
<html lang="en">
<head>
    <title>js login test</title>
    <meta charset="UTF-8">
    <script src="js/jquery.min.js"></script>
    <script type="text/javascript" charset="utf8" src="js/jquery.dataTables.min.js"></script>
    <link rel="stylesheet" href="css/bootstrap.min.css">
    <link rel="stylesheet" href="https://cdn.datatables.net/1.10.19/css/dataTables.bootstrap4.min.css">
</head>
<body>

<form>
    <input type="text" placeholder="Your username" name="username" id="username">
    <button type="button" id="login-btn" name="login-btn">Login</button>
</form>


</body>
<script>
$(document).ready(function() {
   $(document).on('click', '#login-btn', function() {
       var username = $('#username').val();
       if(username === "") {
           alert("No username found");
       }

       $.ajax({
          url: "return.php",
          method: 'POST',
          data:{
            'username': username,
          },
          success: function(data){
            if(data === "success") {
               alert(username);
            }
          },
           error: function () {
               alert("error");
           }
       });
   });
});
</script>
<script src="js/sweetalert.min.js"></script>
</html>

return.php：

<?php

if(isset($_POST)){
    echo "success";
}

?>

Ajax无法获得成功响应

0 个答案: