我有一个Marionette.LayoutView，它调用一个 Backbone.js 集合，获取数据，并根据响应进行渲染。现在我面临的问题是，这个集合需要从两个不同的端点获取数据，这两个端点应该是独立的，然后返回组合的结果。下面是我的代码：
我的Marionette.LayoutView

var View = Marionette.LayoutView.extend({

    template: _.template(some.html),

    regions: {
        div1: '[data-region="div1"]',
        div2: '[data-region="div2"]',
    },

    initialize: function () {
        this.collection = new MovieCollection();
    },

    onRender: function () {
        if (this.collection.length) {
            this.div1.show(new TopMoviesByLikesView({
                collection: this.collection,
                movieCount: 10,
            }));

            this.div2.show(new TopMovieByRatingsView({
                collection: this.collection,
                movieCount: 10,
            }));
        }
    },

});

module.exports = AsyncView.extend({
    ViewConstructor: View,
});

我的Collection

module.exports = Backbone.Collection.extend({

    model: TopMovieModel,

    initialize: function (response) {
        let movieCollection = [];
            let movieSourceOne = new TopMovieFromSourceOne();
            movieSourceOne.fetch({
                success: function (collection, response) {
                    movieCollection = [...movieCollection, ...response.data];
                },
                error: function (collection, response, options) {
                    console.info('~ Response::ERROR', collection, response, options);
                }
            });
        let movieSourceTwo = new movieSourceTwo();
        movieSourceTwo.fetch({
            success: function (collection, response, options) {
                movieCollection = [...movieCollection, ...response.data];
            },
            error: function(collection, response, options) {
                console.info('~ Response::ERROR', collection, response, options);
            }
        });
        this.collection = movieCollection;
    },

我得到的错误是A “url” property or function must be specified有没有方法可以在不使用 Backbone.js 集合中的url的情况下执行此操作？注意：我希望保持两个端点的独立性，因为我不希望在主API失败时收集失败。