BackboneJS传递参数

Question

我有一个主干网应用程序，通过RESTful API从MySQL DB获取数据。现在我想获取与特定artist_id相关的特定视频，但是不管我最终选择了什么艺术家，它都会显示所有的视频，并且在我的控制台中显示../api/videos/artist_videos/undefined --因此，不知怎么的，artist_id没有被传递。

所以我的artistVideo.js看起来是：

function (App, Backbone) {

    var ArtistVideos = App.module();

    ArtistVideos.View = Backbone.View.extend({
        className: 'artistVideos',
        template: 'single_video',
        initialize: function() {
            this.listenTo(this.model, 'sync', this.render);
        },
        serialize: function() {
            return this.model.toJSON();
        },          
        beforeRender: function(){
            $('ul.acmenu li a.selected').removeClass('selected');
            $('ul.acmenu li a.artistVideos').addClass('selected');
        }

    });

    ArtistVideos.ArtistVideosModel = Backbone.Model.extend({
            urlRoot: '../index.php/api/videos/artist_videos/' + this.artist_id,
            defaults: {
                "not categorized": [],
                "live": [],
                "others": []
            }
    });

    return ArtistVideos;
}

然后是我的mainView，artistVideoPage.js，其中包括上面的文件：

define([

'app',
'backbone',
'modules/artistImage',
'modules/artistName',
'modules/artistVideos',
],

function (App, Backbone, Artistimg, Artistname, ArtistVideos ) {

    var ArtistVideo = App.module();

    ArtistVideo.View = Backbone.View.extend({
        className: 'artist',
        template: 'artistVideoPage',
        beforeRender: function() {          
            var artistimgCollection = new Artistimg.ArtistimgCollection();
            artistimgCollection.artist_id = this.artist_id;
            this.insertView('.artistImage', new Artistimg.View({collection: artistimgCollection}));
            artistimgCollection.fetch();

            var artistnameCollection = new Artistname.ArtistnameCollection();
            artistnameCollection.artist_id = this.artist_id;
            this.insertView('.artistName', new Artistname.View({collection: artistnameCollection}));
            artistnameCollection.fetch();

            var artistvideosModel = new ArtistVideos.ArtistVideosModel();
            artistvideosModel.artist_id = this.artist_id;
            this.insertView('.artistVideosDiv', new ArtistVideos.View({model: artistvideosModel}));
            artistvideosModel.fetch();
        }       
    });

    return ArtistVideo;
});

最后，我有一个控制器，如下所示：

ArtistController.prototype.initVideos = function(id) {
    this.artistVideosView.artist_id = id;
    this.artistNavigation.artist_id = id;
    App.useLayout('artistVideoPage', 'artistVideoPage').setViews({
        '.videosDiv': this.artistVideosView,
        '.artistTopMenu': this.artistNavigation
    }).render();
};

我试着做console.log(this.artist_id) -它实际上显示了正确的ID，但我仍然得到了../api/videos/artist_videos/undefined

有人知道这里有什么问题吗？我为什么要得到undefined

提前谢了？

Answer 1

好的，我自己修！

问题是urlRoot --显然它试图马上定义this.artist_id，但它当时并不存在。因此，将其改为：

url: function() {
    return '../index.php/api/videos/artist_videos/' + this.artist_id;
},

解决了问题！