Backbone.Model.save and don't update client

问题:

I want to call save on a Backbone model and have it write data to the server, but not update the client. How do I do this?

To clarify: when Backbone saves a model, it sends all the data to the server, and then retrieves the data from the server and updates it on the client. This second step is what I want not to happen.

To clarify further: The model ('parent' model) has an attribute which is itself a model ('child' model); when it's saved to the server, this child model is converted to JSON. When the parent model updates after the save, the attribute that previously contained a reference to the child model is replaced with the parsed JSON object of the child model that was saved. This is what I need not to happen.

When the data is initially pulled from the server, the parent model "reconstitutes" that object into an appropriate child model, but this is an expensive process and there is no reason to re-do it every time save fires on the parent model, since the child model will never change.

回答1:

It sounds like you do not want to parse your model when you receive the response from the server on a model.save

You can try something such as:

model.save(attributes,{
  success: function() { ... },
  parse  : false     // This will be passed to your parse function as an option
});

You would have to set-up your parse function in your corresponding model as follows:

parse: function(resp,options) {
  // don't update model with the server response if {parse:false} is passed to save
  if (options && !options.parse) return;

  / ... rest of your parse code ... /

Backbone currently defaults options.parse to true. Here is a short-discussion on the topic.

As discussed in that thread, perhaps you want to consider why you do not want want to update the server response to the client. There may be a cleaner way to achieve the results you desire.

回答2:

Depending on how/what your server setup is, all you really have to do is issue a regular AJAX request. This is exactly what backbone does in the background so you'll just bypass the client side logic.

You could do this with native JavaScript, but I'm fairly sure you have some other library in use that can make things much easier.

For the completeness of this answer, I'll give an example with jQuery:

$.ajax({
  type: "POST",
  url: "http://same.as.your.model",
  data: { "the" : "model" },
  dataType: "JSON",
  success: function(){
    // once the request has returned
  }
});

The $.ajax function also has some additional functionality, and you can read about it in the jQuery docs.

回答3:

On client you mean Views? If you want to save your model but not render your views which happens since save will trigger a change event, you should call save with option silent:true, or set a custom option like dontchange:true when calling save and check it in when handling change. I prefer the custom option, because silent has side effects (at least in my version of backbone 1.0.0)

a little code:

when you save:

model.save({},{dontchange: true});

you install your event listeners in the view:

this.listenTo(model, 'change', function(model, options){
  if (options.dontchange)
    return;
  this.render();
});

回答4:

I ran into same problem. model.save(attrs,{patch:true, parse:false}) really did not invoke parse method but model was still merged with server response.

It is not elegant, but this worked for me:

model.clone().save(attrs,{patch:true})

I believe it's best to avoid this situation by clean REST api design.