Question

假设我有模特：

var Book = Backbone.Model.extend({
  // ... 
});

var Store = Backbone.Model.extend({
  getBooks: function(){
    this.books = new Books(App.Singletons.AllBooks.where({storeId: this.id}));
    return this.books;
  }
});

存储has many本书。

还假设我有一个集合：

var Books = Backbone.Collection.extend({
  model: Book
});

在复杂的业务逻辑中，我需要Books的基本商店对象：

var App = App || {};
App.Singletons = App.Singletons || {};
var books = new Books();
books.fetch();
App.Singletons.AllBooks = books;
// ...

为这样的操作同步单例模型的最佳方法是：

var store = new Store({id: 1});
store.fetch();
store.getBooks();

store.books.add(...);
store.books.remove(...);
// etc
// There I need to sync with App.Singletons.AllBooks

现在我在Books个集合中覆盖了这些方法。并且App.Singletons.AllBooks正在同步。我认为应该有另一个更好的解决方案来完成这项任务。谢谢你的帮助。

Answer 1

如果您想要保持“同步”状态，请覆盖同步方法。

在商店的情况下：

var Store = Backbone.Model.extend({
  getBooks: function() {
    this.books = new Books(App.Singletons.AllBooks.where({
      storeId: this.id
    }));
    return this.books;
  },
  sync: function(method, model, options) {
    switch (method) {
      case 'create':
        //your logic to sync with App.Singletons.AllBooks
        break;
      case 'read':
        //your logic to sync with App.Singletons.AllBooks
        break;
      case 'update':
        //your logic to sync with App.Singletons.AllBooks
        break;
      case 'delete':
        //your logic to sync with App.Singletons.AllBooks
        break;
    }
  }
  Backbone.prototype.sync.call(this, method, method, options);
});

嵌套集合的同步

1 个答案: