javascript - 如何使用原生的XMLHttpRequest物件提交表單?

HTML:

<code><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></code>

JavaScript:

<code>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));
    }</code>

form.php

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

程式碼如上, 我想要達到的效果, 是跟在form裡添加了action和method屬性一樣, 提交後可以自動跳到form.pp.
但是這樣提交並沒有反應, 搜了一下, 全是關於jquery的. 還是不知道用ajax提交表單是怎樣的格式.

回覆內容：

HTML:

<code><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></code>

JavaScript:

<code>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));
    }</code>

form.php

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

程式碼如上, 我想要達到的效果, 是跟在form裡添加了action和method屬性一樣, 提交後可以自動跳到form.pp.
但是這樣提交並沒有反應, 搜了一下, 全是關於jquery的. 還是不知道用ajax提交表單是怎樣的格式.

試試將xhr.send(serialize(form))改為

<code>var formData = new FormData(form);
xhr.send(formData);</code>

formData是HTML5用於非同步提交表單的，應該可以滿足樓主的需求。

參考http://www.cnblogs.com/lhb25/...

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