function nicktoid(nick){
if (typeof nick == "String") {
var u = "http://xxxxxx.xxxx/" + nick ;
var xhr2 = new XMLHttpRequest();
xhr2.open('GET',u, true);
xhr2.setRequestHeader('Cache-Control', 'no-cache');
xhr2.setRequestHeader('Pragma', 'no-cache');
xhr2.onreadystatechange = function() {
if (xhr2.readyState == 4) {
txt2 = xhr2.responseText;
var el2 = document.createElement("div");
el2.innerHTML = txt2;
anchors = el2.getElementsByTagName("a");
for (var i=0;i< anchors.length;i++){
if (anchors[i].parentNode.id == "profile_avatar"){
anc.push(anchors[i]);
}
}
var res = anc[0].href;
var arstr = res.split("_");
var resul = arstr[0].substr(6);
}
}
}
}
我有上面的代码。将
var resul
返回到nicktoid例程的正确方法是什么? 最佳答案
使用回调函数,因为Ajax请求异步执行
function nicktoid(nick, callback){
...
xhr2.onreadystatechange = function() {
...
callback && callback(resul);
}
}
nicktoid('foobar', function(resul) {
console.log(resul);
});