Collection Sorting + Updating View about backbone HOT 6 CLOSED

You can use sortBy, as you are, or you can use a comparator function to keep models in a consistent sort order at all times:

http://documentcloud.github.com/backbone/#Collection-comparator

If you change the comparator function, you can call sort() to re-sort the set. If you're listening for "refresh" events, everything should update correctly, automatically.

Hope that helps...

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Thanks - yes that makes sense. So in my case after each refresh, add etc... I would need to clear my view correct?

For example:
If my comparator is sorting models by first name and I'm already displaying Bob and Charlie and than I add Albert, I would need to clear the view and render again.

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after each refresh, add etc... I would need to clear my view correct?

Yes, in the "refresh" event, you would clear and re-render the entire list of models.

If my comparator is sorting models by first name and I'm already
displaying Bob and Charlie and than I add Albert, I would need to
clear the view and render again.

No, you just have to insert new views at the appropriate location in the DOM. Instead of just a simple jQuery .append(), you might want to use insertBefore() or insertAfter(), to put Albert in the right place.

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To do insert before or after - What would be the most efficient way to find it's neighbours in the Collection?

I was thinking if I bind to the add event and get the index of the newly inserted model I could check to see what the next index is or something like that.

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Yep, you can do:

collection.indexOf(model)

... to get its position within the collection. From there, you can look up the adjacent elements, either by id, or by direct reference from the models. For the record, our code that handles this case looks like this:

var view          = new dc.ui.Project({model : model}).render();
var index         = Projects.indexOf(view.model);
var previous      = Projects.at(index - 1);
var previousView  = previous && previous.view;
if (index == 0 || !previous || !previousView) {
  $(this.projectList).prepend(view.el);
} else {
  $(previousView.el).after(view.el);
}

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Very cool - thanks for the help.

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