ajax异步请求步骤
1. 创建 XMLHttpRequest 变量
var ajax = new XMLHttpRequest();
// IE兼容写法 var ajax = new ActiveXObject("Microsoft.XMLHttp");
2. ajax.open()
ajax.open("GET" , url + "?" + data, true) ; 方式 url 是否异步
ajax.open("POST", url, true);
3. ajax.send();
对应GET ajax.send()
对应POST ajax.setRequestHeader("Content-type","appication/x-www-form-urlencode");
ajax.send(data);
4. onreadystatechange
ajax.onreadystatechange = function () {
ajax.readyState == 4 && ajax.status == 200
}