Question

Say I have a Backbone application with a Posts collection. All posts belong to a Blog. Creating a new post requires me to know the Blog it belongs to: POST /blog/42/posts

Here is what I came up with so far — please tell me if there is a better solution:

Knowing that Backbone doesn’t expect me to model the relationships between my models, I simply turned the url attribute into a function to include the Blog ID:

class App.Collections.PostsCollection extends Backbone.Collection
  model: App.Models.Post
  url: -> "/blog/" + this.blogId + "/posts"

(Please excuse the CoffeeScript.) Now I need to make the blogId known to the posts collection. So I’m just tacking it on in the router initialize function:

class App.Routers.PostsRouter extends Backbone.Router
  initialize: (options) ->
    this.posts = new App.Collections.PostsCollection()
    this.posts.blogId = 42  # <----- in reality some meaningful expression ;-)
    this.posts.reset options.positions

That just can’t be right?!

Please enlighten me — how do you usually model these nested collections?

Answer 1

This is one way to solve this problem. An alternative is to create a Blog model. Say you give the Blog model a posts() method. In this case you could use a similar method of attaching the blogId, but in the Blog Model. For example (also in Coffeescript, as above):

class App.Models.Blog extends Backbone.Model
  posts: () =>
    @_posts ?= new App.Collections.PostsCollection({blog: this})

and then in your PostsCollection you would have:

class App.Collections.PostsCollections extends Backbone.Collection
  url: () => "/blog/#{@options.blog.id}/posts"

allowing you to change your router to:

class App.Routers.PostsRouter extends Backbone.Router
  initialize: (options) ->
    # Per your example, an ID of 42 is used below.
    # Ideally, you retrieve the Blog some other way.
    @blog = new App.Models.Blog({id: 42})
    @posts = @blog.posts()

Modifying your structure like this might result in an extra model, but it makes your code much cleaner overall.