Question

我有一个Backbone.View渲染一个集合并在鼠标点击时过滤它。我需要将类active添加到我单击的按钮，但问题是按钮是此视图的一部分，每当我尝试addClass或toggleClass时，它只会再次渲染默认类。这是我的观点：

var ResumeList = Backbone.View.extend({
     events: {
         'click #active': 'showActive',
         'click #passed': 'showPassed'
     },
     initialize: function () {
         this.collection = new ResumeCollection();
     },

     render: function (filtered) {
         var self = this;
         var data;
         if (!filtered) {
             data = this.collection.toArray();
         } else {
             data = filtered.toArray();
         }
         this.$el.html(this.template({ collection: this.collection.toJSON() });

         _.each(data, function (cv) {
             self.$el.append((new ResumeView({model: cv})).render().$el);  
         });
         return this;    
     },

     showActive: function () {
         this.$('#active').toggleClass('active');
         // a function that returns a new filtered collection
         var filtered = this.collection.filterActive();
         this.render(filtered);
     }
});

但正如我已经说过的那样，我需要的类只是暂时切换或添加，然后再次渲染视图并将其设置为默认类。有没有办法解决这个问题？

Answer 1

我简化了渲染并添加了一些优化。

由于我们没有您的模板，因此我将其更改为启用优化：

<button id="active" type="button">Active</button>
<button id="passed" type="button">Passed</button>
<div class="list"></div>

然后你的列表视图可能是这样的：

var ResumeList = Backbone.View.extend({
    events: {
        'click #active': 'showActive',
        'click #passed': 'showPassed'
    },
    initialize: function() {
        this.childViews = [];
        this.collection = new ResumeCollection();
    },

    render: function() {
        this.$el.html(this.template());
        // cache the jQuery element once
        this.elem = {
            $list: this.$('.list'),
            $active: this.$('#active'),
            $passed: this.$('#passed')
        };
        this.renderList(); // default list rendering
        return this;
    },

    renderList: function(collection) {
        this.elem.$list.empty();
        this.removeChildren();
        collection = collection || this.collection.models;

        // Underscore's 'each' has a argument for the context.
        _.each(collection, this.renderItem, this);
    },
    renderItem: function(model) {
        var view = new ResumeView({ model: model });
        this.childViews.push(view);
        this.elem.$list.append(view.render().el);
    },

    showActive: function() {
        this.elem.$active.toggleClass('active');

        var filtered = this.collection.filterActive();
        this.renderList(filtered);
    },

    /**
     * Gracefully call remove for each child view.
     * This is to avoid memory leaks with listeners.
     */
    removeChildren: function() {
        var view;
        while ((view = this.childViews.pop())) {
            view.remove();
        }
    },
});

其他信息：

Managing Views and Memory Leaks
Underscore's each（注意第三个论点）
尽量避免使用callback hell，使回调可重复使用（例如renderItem）

Answer 2

我编辑过这段代码你可以试试吗。

var ResumeList = Backbone.View.extend({
             events: {
                 'click #active': 'filterActive',
                 'click #passed': 'showPassed'
             },

             toggleElement: undefined,

             initialize: function () {
                 this.collection = new ResumeCollection();
             },

             render: function (filtered) {
                 var self = this;
                 var data;
                 if (!filtered) {
                     data = this.collection.toArray();
                 } else {
                     data = filtered.toArray();
                 }
                 this.$el.html(this.template({ collection: this.collection.toJSON() });

                 _.each(data, function (cv) {
                     self.$el.append((new ResumeView({model: cv})).render().$el);  
                 });
                 return this;    
             },

             filterActive: function (evt) {

                 this.toggleElement = this.$el.find(evt.currentTarget);
                 // a function that returns a new filtered collection
                 var filtered = this.collection.filterActive();
                 this.render(filtered);
                 this.toggleActive();
             },

             toggleActive: function() {

                 if(this.toggleElement.is(':checked')) {
                   this.$el.find('#active').addClass('active');
                 } else {
                   this.$el.find('#active').removeClass('active');
                 }
             }
        });

请注意：我已经使用了复选框元素而不是按钮。

在鼠标点击事件

2 个答案: