"How" to save an entire collection in Backbone.js - Backbone.sync or jQuery.ajax?
Solution 1:
My immediate thought is not to override the method on save method on Backbone.Collection but wrap the collection in another Backbone.Model and override the toJSON method on that. Then Backbone.js will treat the model as a single resource and you don't have to hack the way backone thinks too much.
Note that Backbone.Collection has a toJSON method so most of your work is done for you. You just have to proxy the toJSON method of your wrapper Backbone.Model to the Backbone.collection.
var MyCollectionWrapper = Backbone.Model.extend({
url: "/bulkupload",
//something to save?
toJSON: function() {
return this.model.toJSON(); // where model is the collection class YOU defined above
}
});
Solution 2:
A very simple...
Backbone.Collection.prototype.save = function (options) {
Backbone.sync("create", this, options);
};
...will give your collections a save method. Bear in mind this will always post all the collection's models to the server regardless of what has changed. options are just normal jQuery ajax options.
Solution 3:
I ended up just having a 'save' like method and called $.ajax within it. It gave me more control over it without the need to add a wrapper class as @brandgonesurfing suggested (although I absolutely love the idea :) As mentioned since I already had the collection.toJSON() method overridden all I landed up doing was using it in the ajax call...
Hope this helps someone who stumbles upon it...
Solution 4:
This really depends on what the contract is between the client and server. Here's a simplified CoffeeScript example where a PUT to /parent/:parent_id/children
with {"children":[{child1},{child2}]}
will replace a parent's children with what's in the PUT and return {"children":[{child1},{child2}]}
:
class ChildElementCollection extends Backbone.Collection
model: Backbone.Model
initialize: ->
@bind 'add', (model) -> model.set('parent_id', @parent.id)
url: -> "#{@parent.url()}/children" # let's say that @parent.url() == '/parent/1'
save: ->
response = Backbone.sync('update', @, url: @url(), contentType: 'application/json', data: JSON.stringify(children: @toJSON()))
response.done (models) => @reset models.children
return response
This is a pretty simple example, you can do a lot more... it really depends on what state your data's in when save() is executed, what state it needs to be in to ship to the server, and what the server gives back.
If your server is ok with a PUT of [{child1},{child2]
, then your Backbone.sync line could change to response = Backbone.sync('update', @toJSON(), url: @url(), contentType: 'application/json')
.
Solution 5:
The answer depends on what you want to do with the collection on server side.
If you have to send additional data with the post you might need a wrapper model or a relational model.
With the wrapper model you always have to write your own parse method:
var Occupants = Backbone.Collection.extend({
model: Person
});
var House = Backbone.Model.extend({
url: function (){
return "/house/"+this.id;
},
parse: function(response){
response.occupants = new Occupants(response.occupants)
return response;
}
});
Relational models are better I think, because you can configure them easier and you can regulate with the includeInJSON option which attributes to put into the json you send to your rest service.
var House = Backbone.RelationalModel.extend({
url: function (){
return "/house/"+this.id;
},
relations: [
{
type: Backbone.HasMany,
key: 'occupants',
relatedModel: Person,
includeInJSON: ["id"],
reverseRelation: {
key: 'livesIn'
}
}
]
});
If you don't send additional data, you can sync the collection itself. You have to add a save method to your collection (or the collection prototype) in that case:
var Occupants = Backbone.Collection.extend({
url: "/concrete-house/occupants",
model: Person,
save: function (options) {
this.sync("update", this, options);
}
});