jquery中的函数返回undefined

Question

我在jquery中调用的函数返回undefined.我检查了这个函数,当我发现它时它返回了正确的数据.

function addToPlaylist(component_type,add_to_pl_value,pl_list_no) 
    {
        add_to_pl_value_split = add_to_pl_value.split(":");

        $.ajax({
            type: "POST",
            url: "ds/index.php/playlist/check_folder",
            data: "component_type="+component_type+"&value="+add_to_pl_value_split[1],
            success: function(msg)
            {
                if(msg == 'not_folder')
                {
                    if(component_type == 'video')
                    {
                        rendered_item = render_list_item_video(add_to_pl_value_split[0],add_to_pl_value_split[1],pl_list_no)
                    }

                    else if(component_type == 'image')
                    {
                        rendered_item = render_list_item_image(add_to_pl_value_split[0],add_to_pl_value_split[1],pl_list_no)
                    }
                }
                else
                {
                    //List files from folder
                    folder_name = add_to_pl_value_split[1].replace(' ','-');

                    var x = msg; // json 
                    eval('var file='+x); 

                    var rendered_item;

                    for ( var i in file )
                    {
                        //console.log(file[i]);
                        if(component_type == 'video')
                        {
                            rendered_item = render_list_item_video(folder_name+'-'+i,file[i],pl_list_no) + rendered_item;
                        }

                        if(component_type == 'image')
                        {
                            rendered_item = render_list_item_image(folder_name+'-'+i,file[i],pl_list_no) + rendered_item;
                        }
                    }
                }

                $("#files").html(filebrowser_list); //Reload Playlist

                console.log(rendered_item);

                return rendered_item;
            },
            error: function()
            {
                alert("An error occured while updating. Try again in a while");
            }
         })         
    }

$('document').ready(function()
{
    addToPlaylist($('#component_type').val(),ui_item,0); //This one returns undefined
});

Answer 1

该功能addToPlaylist没有return任何意义.它产生一个异步请求,最终执行一个返回函数的回调函数.原始addToPlaylist函数很长时间完成并且在发生这种情况时返回,并且回调函数返回到nobody.

即success: function(msg) { }代码在不同的上下文中执行,并且比周围的addToPlaylist函数晚.

试试这个看看它的实际效果:

function addToPlaylist() {
    $.ajax({
        ...
        success : function () {
            alert('second');  // comes after 'first'
            return null;      // returns to nobody in particular
        }
    });
    alert('first');      // comes before 'second'
    return 'something';  // can only return here to caller
}