刪除集合中的模型並觸發刪除事件 - backbone.js

Question

如何刪除集合中的模型並使remove事件觸發。 我試過people.remove([{ name: "joe3" }]); 但它不會工作。

var Person = Backbone.Model.extend({

    initialize: function () {
        console.log(" person is initialized");
    },
    defaults: {
        name: "underfined",
        age:"underfined"
    }
});

var People = Backbone.Collection.extend({
    initialize: function () {
        console.log("people collection is initialized");
        this.bind('add', this.onModelAdded, this);
        this.bind('remove', this.onModelRemoved, this);
    },
    model: Person,
    onModelAdded: function(model, collection, options) {
        console.log("options = ", options);
        alert("added");
    },
    onModelRemoved: function (model, collection, options) {
        console.log("options = ", options);
        alert("removed");
    },
});

//var person = new Person({ name: "joe1" });
var people = new People();



//people.add([{ name: "joe2" }]);
people.add([{ name: "joe1" }]);
people.add([{ name: "joe2" }]);
people.add([{ name: "joe3" }]);
people.add([{ name: "joe4" }]);
people.add([{ name: "joe5" }]);

people.remove([{ name: "joe3" }]);



console.log(people.toJSON());

Answer 1

對於其他尋找刪除位置的人，您可以使用collection.where調用鏈接它。 像這樣刪除與搜索匹配的所有項目：

people.remove(people.where({name: "joe3"}));

請參閱Backbone collection.where

Answer 2

通過做：

people.remove([{ name: "joe3" }]);

您不刪除模型，因為您只傳遞一個未連接到people集合的普通對象。 相反，你可以做這樣的事情：

people.remove(people.at(2));

要么：

var model = new Person({name: "joe3"});
people.add(model);
...
people.remove(model);

也會奏效。

所以你需要從集合中引用實際的模型對象;

http://jsfiddle.net/kD9Xu/

Answer 3

var Person = Backbone.Model.extend({
    defaults: {
        name: "underfined",
        age:"underfined"
    }
});

var People = Backbone.Collection.extend({
    initialize: function () {
        this.bind('remove', this.onModelRemoved, this);
    },
    model: Person,
    onModelRemoved: function (model, collection, options) {
        alert("removed");
    },
    getByName: function(name){
       return this.filter(function(val) {
          return val.get("name") === name;
        })
    }
});

var people = new People();

people.add(new Person({name:"joe1"}));
people.add(new Person({name:"joe2"}));
people.remove(people.getByName("joe1"));

console.info(people.toJSON());

Answer 4

另一種方法是縮短一點，並為集合觸發remove事件：

people.at(2).destroy();
// OR
people.where({name: "joe2"})[0].destroy();

觸發模型上的“destroy”事件，該事件將通過包含它的任何集合冒泡.http：//backbonejs.org/#Model-destroy

Answer 5

要刪除“[0]”，您可以使用以下代碼：

people.findWhere({name: "joe2"}).destroy();