ajax - 检索跨浏览器 XmlHttpRequest 的最简单方法

Question

检索适用于所有浏览器的 XmlHttpRequest 对象的最简单、最安全的方法是什么？无需任何额外的库。有没有您经常使用的代码片段？

附注我知道网上有大量的例子，但这正是我要问的原因:有太多不同的例子，我只想要一些简单且经证明有效的东西。

jQuery 和其他库不是一个选项。 Why does jquery leak memory so badly?

Answer 1

虽然我建议使用完整的库来简化使用，但在现代浏览器中发出 AJAX 请求可能相当简单:

var req = new XMLHttpRequest();
req.onreadystatechange = function(){
    if(this.readyState == 4){
        alert('Status code: ' + this.status);
        // The response content is in this.responseText
    }
}
req.open('GET', '/some-url', true);
req.send();

以下代码片段是基于 quirksmode.org 中的代码片段的更高级代码片段甚至支持非常旧的浏览器(早于 Internet Explorer 7):

function sendRequest(url,callback,postData) {
    var req = createXMLHTTPObject();
    if (!req) return;
    var method = (postData) ? "POST" : "GET";
    req.open(method,url,true);
    // Setting the user agent is not allowed in most modern browsers It was
    // a requirement for some Internet Explorer versions a long time ago.
    // There is no need for this header if you use Internet Explorer 7 or
    // above (or any other browser)
    // req.setRequestHeader('User-Agent','XMLHTTP/1.0');
    if (postData)
        req.setRequestHeader('Content-type','application/x-www-form-urlencoded');
    req.onreadystatechange = function () {
        if (req.readyState != 4) return;
        if (req.status != 200 && req.status != 304) {
//          alert('HTTP error ' + req.status);
            return;
        }
        callback(req);
    }
    if (req.readyState == 4) return;
    req.send(postData);
}

var XMLHttpFactories = [
    function () {return new XMLHttpRequest()},
    function () {return new ActiveXObject("Msxml3.XMLHTTP")},
    function () {return new ActiveXObject("Msxml2.XMLHTTP.6.0")},
    function () {return new ActiveXObject("Msxml2.XMLHTTP.3.0")},
    function () {return new ActiveXObject("Msxml2.XMLHTTP")},
    function () {return new ActiveXObject("Microsoft.XMLHTTP")}
];

function createXMLHTTPObject() {
    var xmlhttp = false;
    for (var i=0;i<XMLHttpFactories.length;i++) {
        try {
            xmlhttp = XMLHttpFactories[i]();
        }
        catch (e) {
            continue;
        }
        break;
    }
    return xmlhttp;
}