suppose I have a model and a view ,ths view have two method:one is bind the document mousemove event and the other is unbind method,defalut I give the document mousemove event, once the model’s enable value changed I will call the view’s unbind method:
window.ConfigModel = Backbone.Model.extend({
defaults: {
'enable':0
},
initialize: function(){
this.bind("change:enable", function () {
var portView2 = new PortView();
portView2.viewOff();
});
},
change:function () {
this.set('enable', 9);
}
})
window.PortView = Backbone.View.extend({
viewOn: function () {
$(document).on('mousemove', function () {
console.log('move')
})
},
viewOff: function () {
$(document).off('mousemove');
}
})
then I put an input on the document to call the model changed:
$('input').click(function () {
var configModel = new ConfigModel();
configModel.change();
})
the boot script is :
var portView1 = new PortView();
portView1.viewOn();
The problem is once I call the click the input button ,the chrome would tell me an error:Maximum call stack size exceeded it seems the change be invoke many times.So what’s the problem with my problem ,how can I solve this problem
Backbone models already have a
changemethod:Presumably something inside Backbone is trying to call
configModel.change()and getting your version ofchangewhich triggers anotherchange()call inside Backbone which runs yourchangewhich … until the stack blows up.You should use a different name for your
changemethod.That said, your code structure is somewhat bizarre. A model listening to events on itself is well and good but a model creating a view is odd:
And instantiating a view simply to call a single method and then throw it away is strange as is creating a new model just to trigger an event.
I think you probably want to have a single ConfigModel instance as part of your application state, say
app.config. Then yourclickhandler would talk to that model:Then you’d have some other part of your application (not necessarily a view) that listens for changes to
app.configand acts appropriately:And then start the application with a single
app.init()call: