闲来修复本站全局播放器(APlayer)

这是原始代码:

function ap_list () {
    var a = [];
    $.ajax({
        url: "https://api.******.com/******/api?server=*******&type=playlist&id=**********",
        async :0,
        success: function(t) {
            a = JSON.parse(t);
        }
    });
    $.ajax({
        url: "https://api.******.com/******/api?server=*******&type=song&id=********",
        async :0,
        success: function(t) {
            a = a.concat(JSON.parse(t));
        }
    });
    $.ajax({
        url: "https://api.*******.com/******/api?server=*******&type=song&id=*********",
        async :0,
        success: function(t) {
            a = a.concat(JSON.parse(t));
        }
    });
    return a;
}
function aplayer_fixed () {
    var a = new APlayer({
        element: document.getElementById("aplayer-fixed"),
        mutex: !0,
        theme: "#97dffd",
        order: "random",
        lrcType: 3,
        fixed: !0
    });
    a.list.add(ap_list());
}
aplayer_fixed();

由于我使用Hack方法手动拼接了一个播放列表(一个歌单+两首单曲),于是便无法使用异步ajax请求,只好把async :0设为异步请求,但是一直有一个黄色的警告⚠提醒:

很不爽...

今天一拍脑门,突然想到了jQuery的when().done()方法.

于是对代码稍作修改:

(function () {
    var playlist=[];
    $.when(
    $.getJSON("https://api.******.com/******/api?server=*******&type=playlist&id=**********"),
    $.getJSON("https://api.******.com/******/api?server=*******&type=song&id=********"),
    $.getJSON("https://api.******.com/******/api?server=*******&type=song&id=*********")
    ).then(function(a, b, c) {
        playlist = a[0].concat(b[0], c[0]);
        var a = new APlayer({
            element: document.getElementById("aplayer-fixed"),
            mutex: !0,
            theme: "#97dffd",
            order: "random",
            lrcType: 3,
            fixed: !0
        });
        a.list.add(playlist);
    });
})();

其中为了避免playlist的变量名对全局产生污染,作了闭包处理.

目前再也没有黄色的警告⚠提醒啦!