封装自己的ajax函数

function MyAjax(url,onsuccess,onfail) {
    var xhr = window.XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");
    xhr.open("POST", url, true);
    xhr.onreadystatechange = function () {
        if (xhr.readyState == 4) {
            if (xhr.status == 200) {
                onsuccess(xhr.responseText);
            }
            else {
                onfail(xhr.status);
            }
        }
    };
    xhr.send();
}