如何使用原生的XMLHttpRequest对象提交表单?

Question

HTML:

<form id="info">
    <label for="id">ID: </label>
    <input type="text" name="id">
    <br>
    <label for="user">User: </label>
    <input type="text" name="user" id="user">
    <br>
    <input type="submit" name="submit" id="submit">
</form>

JavaScript:

var submit = document.getElementById("submit");
        submit.onclick = function() {
            var xhr = new XMLHttpRequest();
            xhr.onreadystatechange = function(){
                if(xhr.state == 4) {
                    if((xhr.status >= 200 && xhr.status < 300) || xhr.status == 304) {
                        console.log(xhr.responseText);
                    } else {
                        alert("HttpRequest was unsccessful: " + xhr.status);
                    }
                }
            }
            xhr.open("post", "form.php", true);
            xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
            var form = document.getElementById("info");
            xhr.send(serialize(form));
    }

form.php

<?php 
$id = $_POST['id'];
$user = $_POST['user'];
echo $id . $user;
?>

代码如上, 我想要达到的效果, 是跟在form里添加了action和method属性一样, 提交后可以自动跳转到form.pp.
但是这样提交并没有反应, 搜了一下, 全是关于jquery的. 还是不知道用ajax提交表单是怎样的格式.

Answer 1

试试将xhr.send(serialize(form))改为

var formData = new FormData(form);
xhr.send(formData);

formData是HTML5用于异步提交表单的，应该可以满足楼主的需求。

参考http://www.cnblogs.com/lhb25/...

Answer 2

var fd = new FormData;
fd.append("username",document.querySelector("#username").value);
fd.append("password",document.querySelector("#password").value);
...
xhr.send(fd);