Question

js的ajax代码，不用任何框架，就用js怎么实现ajax？

Answer 1

var xmlhttp;
if (window.XMLHttpRequest) 
    xmlhttp=new XMLHttpRequest(); 
else 
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
//创建xmlhttp对象
xmlhttp.onreadystatechange=function()
{
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
     {
        // 成功后执行的代码
        //服务器返回数据为xmlhttp.responseText
     }
   
}
xmlhttp.open(方法,url,是否异步);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send(data);
//get时data为null

Answer 2

<script type="text/javascript">
// 封装创建XMLHttpRequest对象
function createXMLHTTPObject() {
var XMLHttpFactories = [
function () { return new XMLHttpRequest() },
function () { return new ActiveXObject("Msxml2.XMLHTTP") },
function () { return new ActiveXObject("Msxml3.XMLHTTP") },
function () { return new ActiveXObject("Microsoft.XMLHTTP") },
];
var xmlhttp = false;
for (var i = 0; i < XMLHttpFactories.length; i++) {
try {
xmlhttp = XMLHttpFactories[i]();
}
catch (e) {
continue;
}
break;
}
return xmlhttp;
}
//封装异步请求函数
function request(url, callback, data) {
var xmlHttp = createXMLHTTPObject();
if (!xmlHttp) return;
var method = (data) ? "POST" : "GET";
xmlHttp.open(method, url, true);
xmlHttp.setRequestHeader('User-Agent', 'XMLHTTP/1.0');
if (data)
xmlHttp.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
xmlHttp.onreadystatechange = function () {
if (xmlHttp.readyState != 4) return;
if (xmlHttp.status != 200 && xmlHttp.status != 304) {
alert('HTTP请求错误 ' + xmlHttp.status);
return;
}
callback(xmlHttp);
}
if (xmlHttp.readyState == 4) return;
xmlHttp.send(data);
}
</script>

<script type="text/javascript">
//定义请求地址
var mi = document.domain
var url = "http://ip-api.com/json/?lang=zh-CN";
//定义回调函数，参数为XMLHttpRequest对象
var callback = function (xmlHttp) {
var info = xmlHttp.responseText;
var o = eval("(" + info + ")");

$("#michen").html("通过接口动态查询IP" + JSON.stringify(o));

//alert(info);
// alert(o[1].name);

}
//调用请求函数
request(url, callback);

</script>
<span id="michen"></span>