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我刚开始使用 Backbone。根据下面的代码，我有两个问题要问。

第一个问题是在我填写表格并单击按钮后，应该使用一些默认属性创建模型对象。但是，console.log 会在我将模型传递给新视图对象之前使用我填写的表单中的最新属性打印模型。

第二个问题是我可以成功地将数据保存到db，但是我的成功回调函数没有被调用。有人可以帮我回答这些问题吗？

var form = document.forms[0];

var RetailerModel = Backbone.Model.extend({
    urlRoot: ' retailer.php',
    defaults: {
        name: 'company-name',
        address: 'company-address',
        phone: 'company-phone',
        icon: 'http://localhost/icon.png'
    }
});

var RetailerCollection = Backbone.Collection.extend({

});

var RetailerView = Backbone.View.extend({

    className: 'retailer',

    template: _.template($('#retailer-template').html()),

    initialize: function() {
        //this.listenTo(this.model, 'change', this.render);

        var obj = {
            name: form.name.value,
            address: form.address.value,
            phone: form.phone.value
        };

        this.model.set(obj);
        //why the successful callback does not work????
        this.model.save(null, {success: function(model, response){console.log('successful');}});
    },

    render: function() {
        $('#retailer-list').append(this.$el.html(this.template(this.model.toJSON())));

        return this;
    }
});

var RetailerViews = Backbone.View.extend({

});

$('#submit').click(function(e){
    var retailer_model = new RetailerModel();
    console.log(retailer_model); // this prints out the new changed attributes instead of the default ones, why???
    var retailer_view = new RetailerView({model: retailer_model});
    form.reset();
});
4

2 回答 2

3

1 获取模型的正确状态

console.log(retailer_model)不会显示模型属性，它会显示整个模型，但console.log(retailer_model.attributes)会。还要记住，这console.log 并不总是正确的，特别是如果您在记录后修改对象，这可能会导致混乱！

要获得模型的实际当前状态，您应该对其进行浅拷贝或深拷贝。因此，console.log(model)您可以使用下划线克隆方法进行浅拷贝：

console.log(_(model).clone());

2 保存成功回调

为了提供帮助，我们真的需要更多地了解您的情况。我要寻找的第一件事是您的服务器是否正在返回正确的反馈。Backbone 是一个基于 REST 标准的系统。您确定您的服务器正在返回正确的响应代码吗？如果您使用 chrome，请打开您的开发人员工具，然后打开网络选项卡以检查您在发布模型时得到的响应。为了触发成功回调，服务器应该返回 200 或 201 的状态。

另一种测试方法是查看错误回调是否正在触发:)。

于 2013-04-27T21:14:30.483 回答
1

I tried to see your errors, I realized that you missed these things:
1. You should do DOM related things when DOM ready event fired:

$(function() { 
    form = document.forms[0];
    $(form).live('submit', function(e){
        // ...
    });
});


2. You can use submit for forms. It catches also enter click within form: 

$(form).live('submit',function(){
    //...
});


3. You should use return false; within submit form. It prevents default form data sending to action url.

$(form).live('submit', function(e){
    // ...
    return false;
});

So, it looks like this. I didn't check success callback, but I hope it will work. 

<html>
<head>
<script src="jquery-1.7.2.js" type="text/javascript"></script>
<script type="text/javascript" src="underscore.js"></script>
<script type="text/javascript" src="backbone.js"></script>
<script type="text/javascript">

var RetailerModel = Backbone.Model.extend({
    urlRoot: ' retailer.php',
    defaults: {
        name: 'company-name',
        address: 'company-address',
        phone: 'company-phone',
        icon: 'http://localhost/icon.png'
    }
});

var RetailerCollection = Backbone.Collection.extend({});

var RetailerView = Backbone.View.extend({

    className: 'retailer',

    //template: _.template($('#retailer-template').html()),

    initialize: function() {
        //this.listenTo(this.model, 'change', this.render);

        var obj = {
            name: form.name.value,
            address: form.address.value,
            phone: form.phone.value
        };

        this.model.set(obj);
        //why the successful callback does not work????
        this.model.save(null, {
            success: function(model, response){
                console.log('successful');
            }
        });
    },

    render: function() {
        $('#retailer-list').append(this.$el.html(this.template(this.model.toJSON())));
        return this;
    }
});

var RetailerViews = Backbone.View.extend({});
$(function() { // you should do DOM related things when DOM ready event fired
    form = document.forms[0];
    $(form).live('submit', function(e){
        var retailer_model = new RetailerModel();
        console.log(retailer_model);
        var retailer_view = new RetailerView({model: retailer_model});
        form.reset();

        return false; // to prevent form data sending return false
    });
});
</script>

</head>
<body>
    <form action="#" id="#submit">
        <input type="submit" value="submit"/>
        <input type="text" name="address"/>
        <input type="text" name="name"/>
        <input type="text" name="phone"/>
    </form>
</body>
</html>
于 2013-04-27T21:51:24.387 回答