XMLHttpRequest 发布 HTML 表单 [英] XMLHttpRequest to Post HTML Form

问题描述

当前设置

我有一个这样的 HTML 表单.

I have an HTML form like so.

<form id="demo-form" action="post-handler.php" method="POST">
   <input type="text" name="name" value="previousValue"/>
   <button type="submit" name="action" value="dosomething">Update</button>
</form>

我可能在一个页面上有很多这样的表单.

I may have many of these forms on a page.

我的问题

如何异步提交此表单而不被重定向或刷新页面?我知道如何使用 XMLHttpRequest.我遇到的问题是从 javascript 中的 HTML 中检索数据，然后将其放入发布请求字符串中.这是我目前用于 zXMLHttpRequest` 的方法.

How do I submit this form asynchronously and not get redirected or refresh the page? I know how to use XMLHttpRequest. The issue I have is retrieving the data from the HTML in javascript to then put into a post request string. Here is the method I'm currently using for my zXMLHttpRequest`'s.

function getHttpRequest() {
    var xmlhttp;
    if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    } else {// code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    
    return xmlhttp;
}

function demoRequest() {
       var request = getHttpRequest();
       request.onreadystatechange=function() {
             if (request.readyState == 4 && request.status == 200) {
                   console.log("Response Received");
             }
       }
       request.open("POST","post-handler.php",true);
       request.setRequestHeader("Content-type","application/x-www-form-urlencoded");
       request.send("action=dosomething");
}

例如，假设在单击表单的提交按钮时调用了 javascript 方法 demoRequest()，我如何从该方法访问表单的值，然后将其添加到 XMLHttpRequest?

So for example, say the javascript method demoRequest() was called when the form's submit button was clicked, how do I access the form's values from this method to then add it to the XMLHttpRequest?

编辑

试图从下面的答案中实现解决方案，我已经像这样修改了我的表单.

Trying to implement a solution from an answer below I have modified my form like so.

<form id="demo-form">
       <input type="text" name="name" value="previousValue"/>
       <button type="submit" name="action" value="dosomething" onClick="demoRequest()">Update</button>
</form>

但是，在单击按钮时，它仍然试图将我重定向(到我不确定的地方)并且我的方法没有被调用?

However, on clicking the button, it's still trying to redirect me (to where I'm unsure) and my method isn't called?

按钮事件监听器

document.getElementById('updateBtn').addEventListener('click', function (evt) {
                                evt.preventDefault();
                                
                                // Do something
                                updateProperties();
                                
                                return false;
                            });

推荐答案

POST 字符串格式如下:

The POST string format is the following:

name=value&name2=value2&name3=value3

因此，您必须获取所有名称、它们的值并将它们放入该格式.您可以迭代所有输入元素或通过调用 document.getElementById() 来获取特定元素.

警告:您必须使用encodeURIComponent() 用于所有名称，尤其是值，以便字符串中可能包含的 & 不会破坏格式.

Warning: You have to use encodeURIComponent() for all names and especially for the values so that possible & contained in the strings do not break the format.

示例:

var input = document.getElementById("my-input-id");
var inputData = encodeURIComponent(input.value);

request.send("action=dosomething&" + input.name + "=" + inputData);

---

另一个更简单的选择是使用 FormData 对象.这样的对象可以保存名称和值对.

---

Another far simpler option would be to use FormData objects. Such an object can hold name and value pairs.

幸运的是，我们可以从现有表单构造一个 FormData 对象，然后将其直接发送到 XMLHttpRequest 的方法 send():

Luckily, we can construct a FormData object from an existing form and we can send it it directly to XMLHttpRequest's method send():

var formData = new FormData( document.getElementById("my-form-id") );
xhr.send(formData);

这篇关于XMLHttpRequest 发布 HTML 表单的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！