无法从函数内部获取 return 值

Question

这个函数不允许我在函数外传递数据。它 return 是正确的数据，但每当我尝试将它添加到数组或函数外部的类似内容时，它不会 return 任何东西。我知道一些多余的 json 解析和字符串化，但请忽略它。我试过设置全局变量，但没有成功。

var item.jid = 342323; //example
p = connection.vCard.get(item.jid);   

p.done(function(vCard) {
    var json = JSON.stringify(vCard);
    var json = $.parseJSON(json);

    $(json).each(function(item,val) {
        $.each(val,function(k,v) {
            var json2 = JSON.stringify(v);
            var obj = JSON.parse(json2);
            var what = obj.BINVAL;

            if(what != undefined) {
                var bgimg = what;   
                bgimg = "data:image/jpeg;base64,"+bgimg;
                //console.log(bgimg); //This returns correct data, but only from within this each loop.
                //return bgimg; //this does not return anything
                //arr.push(bgimg);//this does not push anything to array
                //console.log(item.jid); //does not return global var
            }
        });
    });           
});

编辑
这是我试图提取的函数（根据 romy 的回答进行了更改）以进一步说明：我正在构建一个带有化身的联系人名单。

var bgimg=null;
var arr = new Array();
function getAvatar(vCard) {
            var json = JSON.stringify(vCard);
            var json = $.parseJSON(json);
                $(json).each(function(item,val){
                    $.each(val,function(k,v){
                        var json2 = JSON.stringify(v);
                        var obj = JSON.parse(json2);
                        var what = obj.BINVAL;
                            if(what != undefined){
                                bgimg = what;   
                                bgimg = "data:image/jpeg;base64,"+bgimg;
                                arr.push(bgimg);
                                //console.log(bgimg); //This returns correct data.
                            }

                    }); 
                });
         };

function getRoster(){

connection.roster.requestRoster(function(roster){
   var items = roster.getItems();
   for( var i = 0; i < items.length; i++ ){
        var arr = new Array();
        var item = items[i];
        var thejid = item.jid;
        var thename = item.name;
        arr.push(thejid);
        arr.push(thename);
        p = $.Deferred();
        p = connection.vCard.get(item.jid);   
            p.done(function(vCard) {
                getAvatar(vCard);
            });
       p.resolve();
       console.log(arr);
var theavatar = //This would return whatever the deferred function would throw me. 

        $('.sidebar-nav').append("<li class='contacts'><div id='"+thejid+"' class='avatar' src='"+theavatar+"'></div><a href='#' class='"+thejid+"'>"+thename+"</a></li>");

   }

Answer 1

您处于 async/deferred 执行模式。

我看不出像您那样对 return 有任何好处（它也不会起作用）：

You can use global variable if you want :

var d =  $.Deferred();

var myResult=null;

d.done(function (a){ myResult=a;}) .done(function (){ alert(window["myResult"])})


d.resolve(1); //will alert "1"

但我更愿意：

var d =  $.Deferred();


function doWhatever(x)
{
  //do wtf you want
  alert(x)
}

var myResult=null;

d.done(function (a){ doWhatever(a)});


d.resolve(1);

http://jsbin.com/potula/1/edit

无法从函数内部获取 return 值

Unable to return value from inside a function

variables

jquery

return-value