Ajax 表单提交在第一次提交时返回错误，但在第二次单击时提交
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【中文标题】Ajax 表单提交在第一次提交时返回错误，但在第二次单击时提交【英文标题】：Ajax form submit returns error on first submit but submits on second click 【发布时间】：2017-07-27 03:42:34 【问题描述】：我正在尝试使用 ajax 表单提交来提交表单而不刷新页面。但是，当我第一次单击提交按钮时，它会警告错误（在警告中它显示了我有提交表单的整个 html 页面）。如果我关闭警报并再次点击相同的提交按钮而不刷新页面，它会完美地提交表单。
因此，每次刷新页面时，我都必须点击提交按钮，然后关闭错误警报，然后再次提交。我很确定它在 ajax 文件中有一些语法错误。
$('document').ready(function()
/* validation */
$("#login-form").validate(
rules:
upass:
required: true
,
uemail:
required: true,
email: true
,
,
messages:
upass:
required: "please enter your password"
,
uemail: "please enter your email address",
,
errorPlacement: function(error, element)
$(element).closest('.form-group').find('.help-block').html(error.html());
,
highlight: function(element)
$(element).closest('.form-group').removeClass('has-success').addClass('has-error');
,
submitHandler: submitForm
);
/* validation */
/* login submit */
function submitForm()
var data = $("#login-form").serialize();
$.ajax(
type: 'POST',
url: 'login.php',
data: data,
dataType: 'json',
success: function(data)
// console.log(data);
// alert(JSON.stringify(response));
// if(response == 1)
if (data.status == 'success')
// alert('Good!');
alert(JSON.stringify(data));
window.location.href = "home.php";
else
window.location.href = "login_failed.php";
,
error: function(data)
// console.log(JSON.stringify(data));
alert(JSON.stringify(data));
);
return false;
);HTML 文件
<!DOCTYPE HTML>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<script src="js/jquery.validate.min.js"></script>
<!-- Latest compiled and minified javascript -->
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js" integrity="sha384-Tc5IQib027qvyjSMfHjOMaLkfuWVxZxUPnCJA7l2mCWNIpG9mGCD8wGNIcPD7Txa" crossorigin="anonymous"></script>
<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" integrity="sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u" crossorigin="anonymous">
<!-- Optional theme -->
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap-theme.min.css" integrity="sha384-rHyoN1iRsVXV4nD0JutlnGaslCJuC7uwjduW9SVrLvRYooPp2bWYgmgJQIXwl/Sp" crossorigin="anonymous">
<link rel="stylesheet" href="assets/signup-form.css" type="text/css" />
</head>
<body>
<!--container starts-->
<div class="container">
<!--Header starts-->
<nav class="navbar navbar-fixed-top">
<div class="container-fluid">
<div class="logo">
<a href="index.php"><img src="images/logo.png" style="float:left"/></a>
</div>
<form method="post" action="login.php" id="login-form" class="form-inline" autocomplete="off">
<div class="form-group"><strong style="color:white; font-family:tahoma">Email:</strong>
<input type="email" class="form-control" placeholder="Email address" name="uemail" id="user_email" />
</div>
<!--<input type="email" name="uemail" placeholder="Email" required/>-->
<!--<input type="email" name="u_email" placeholder="Email" required="required"/>-->
<div class="form-group"><strong style="color:white; font-family:tahoma">Password:</strong>
<input type="password" class="form-control" placeholder="Password" name="upass" />
</div>
<!--<input type="password" name="upass" placeholder="Password" required/>-->
<!--<input type="password" name="u_pass" placeholder="********" required="required"/>-->
<button type="submit" name="login" class="btn btn-default">Login</button></br>
<!--<button type="submit" name="login"> Login </button>-->
<a href="forgot_password.php">Forgot your password?</a>
</form>
</div>
</nav>
<script src="js_files/login.js"></script>
<!-- <script src="js_files/login_failed.js"></script> -->
<!-- <script src="register_script.js"></script> -->
<script src="js_files/sign_up.js"></script>
</body>
</html>
这是 php 文件 login.php
<?php
require_once 'class/head.class.php';
if(isset($_POST['login']))
$u_email = $_POST['uemail'];
$u_pass = $_POST['upass'];
try
if(init::loginCheck($u_email) < 3)
if(users::login($u_email, $u_pass))
init::loginSuccess($u_email);
// init::redirect('home.php');
$response['status'] = 'success';
// $response['message'] = '<div class="alert alert-danger"><span class="glyphicon glyphicon-ok"></span> registered sucessfully, please confirm your email</div>';
// echo 1;
else
init::loginAttempt($u_email);
// init::redirect('home.php');
$response['status'] = 'redirect'; // could not register
// $response['message'] = '<div class="alert alert-warning"><span class="glyphicon glyphicon-info-sign"></span> You have entered incorrect login details. Please try again</div>';
// echo 0;
else
$response['status'] = 'redirect';
catch(PDOException $e)
echo $e->getMessage();
echo json_encode($response);
?>
它返回的错误是包含表单的html页面：
"readyState":4,"responseText":"\n\n \n\n\n\nhttps://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min。 js\">\n \n\n\nhttps://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js\" 完整性=\"sha384-Tc5IQib027qvyjSMfHjOMaLkfuWVxZxUPnCJA7l2mCWNIpG9mGCD8wGNIcPD7Txa\" crossorigin=\"匿名\">\n\n\n\nhttps://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css\" 完整性=\"sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u\" crossorigin= \"匿名\">\n\n\nhttps://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap-theme.min.css\" 完整性=\"sha384-rHyoN1iRsVXV4nD0JutlnGaslCJuC7uwjduW9SVrLvRYooPp2bWYgmgJQIXwl/Sp\" 跨域=\"匿名\">\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n\n\n \n \n电子邮件：\ n \n \n \n\n\n \n -->\n -->\n\n 密码：\n \n\n \n\n \n -->\n -->\n 登录\n 登录 -->\n\n 忘记密码？\n \n \n \n\n \n\n \n -->\n -->\n \n\n \n","status":200,"statusText":"OK"
【问题讨论】：
有我们可以看到的任何 html 吗？ 它返回什么错误？ 表格的代码.. @Rick 。我已经添加了 html 和 php 代码。谢谢 @santoshgore 。我添加了它显示的错误。谢谢 【参考方案1】：将此添加到您的 ajax 请求中
async: false,
这将修复您的错误并添加它以查看来自 ajax 请求的任何错误
error: function(er, err, error)
alert(error);
;
【讨论】：
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