Question

我正在尝试使用jQuery捕获提交事件，然后将格式化为JSON的表单元素发送到PHP页面。我在捕获提交时遇到问题，我从.click()事件开始，但转而转移到.submit()。

我现在有以下修剪过的代码。

HTML

<form method="POST" id="login_form">
    <label>Username:</label>
    <input type="text" name="username" id="username"/>
    <label>Password:</label>
    <input type="password" name="password" id="password"/>
    <input type="submit" value="Submit" name="submit" class="submit" id="submit" />
</form>

的Javascript

$('#login_form').submit(function() {
    var data = $("#login_form :input").serializeArray();
    alert('Handler for .submit() called.');
});

Answer 1

将文档中的代码包装好并阻止默认的提交操作：

$(function() { //shorthand document.ready function
    $('#login_form').on('submit', function(e) { //use on if jQuery 1.7+
        e.preventDefault();  //prevent form from submitting
        var data = $("#login_form :input").serializeArray();
        console.log(data); //use the console for debugging, F12 in Chrome, not alerts
    });
});

Answer 2

试试这个：

使用'return false'来减少事件的流程：

$('#login_form').submit(function() {
    var data = $("#login_form :input").serializeArray();
    alert('Handler for .submit() called.');
    return false;  // <- cancel event
});

修改

确认表单元素是否具有jQuery的'length'：

alert($('#login_form').length) // if is == 0, not found form $('#login_form').submit(function() { var data = $("#login_form :input").serializeArray(); alert('Handler for .submit() called.'); return false; // <- cancel event });

OR：

等待DOM准备就绪：

jQuery(function() { alert($('#login_form').length) // if is == 0, not found form $('#login_form').submit(function() { var data = $("#login_form :input").serializeArray(); alert('Handler for .submit() called.'); return false; // <- cancel event }); });

您是否将代码放入事件“准备好”文档中或在DOM准备好之后？

Answer 3

只需将form.submit函数替换为您自己的实现：

var form = document.getElementById('form');
var formSubmit = form.submit; //save reference to original submit function

form.onsubmit = function(e)
{
    formHandler();
    return false;
};

var formHandler = form.submit = function()
{
    alert('hi there');
    formSubmit(); //optionally submit the form
};

Answer 4

只是一个提示：记得将代码检测放在document.ready上，否则可能无效。那是我的理由。

Answer 5

$(document).ready(function () {
  var form = $('#login_form')[0];
  form.onsubmit = function(e){
  var data = $("#login_form :input").serializeArray();
  console.log(data);
  $.ajax({
  url: "the url to post",
  data: data,
  processData: false,
  contentType: false,
  type: 'POST',
  success: function(data){
    alert(data);
  },
  error: function(xhrRequest, status, error) {
    alert(JSON.stringify(xhrRequest));
  }
});
    return false;
  }
});

<!DOCTYPE html>
<html>
<head>
<title>Capturing sumit action</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<form method="POST" id="login_form">
    <label>Username:</label>
    <input type="text" name="username" id="username"/>
    <label>Password:</label>
    <input type="password" name="password" id="password"/>
    <input type="submit" value="Submit" name="submit" class="submit" id="submit" />
</form>

</body>

</html>

使用jquery和.submit捕获表单提交

5 个答案: