同步时排除模特属性（Backbone.js的） [英] Exclude model properties when syncing (Backbone.js)

问题描述

有没有办法从我的模型排除某些属性时，我同步？

Is there a way to exclude certain property from my model when I sync?

例如，我请一些视图状态我的模型信息。比方说，我有一个选择器模块，并且该模块只需拨动一个选我的模型属性。后来，当我打电话 .save（）我的收藏，我愿意忽略的选择的值和从同步到服务器中排除。

For example, I keep in my model information about some view state. Let's say I have a picker module and this module just toggle a selected attributes on my model. Later, when I call .save() on my collection, I'd want to ignore the value of selected and exclude it from the sync to the server.

是否有这样做的一个干净的方式？

Is there a clean way of doing so?

（让我知道，如果你想了解更多详细信息的）

推荐答案

这似乎是最好的解决方案（基于@nikoshr引用的问题）

This seems like the best solution (based on @nikoshr referenced question)

Backbone.Model.extend({

    // Overwrite save function
    save: function(attrs, options) {
        options || (options = {});
        attrs || (attrs = _.clone(this.attributes));

        // Filter the data to send to the server
        delete attrs.selected;
        delete attrs.dontSync;

        options.data = JSON.stringify(attrs);

        // Proxy the call to the original save function
        return Backbone.Model.prototype.save.call(this, attrs, options);
    }
});

所以，我们覆盖节省模型实例功能，但我们只是过滤掉我们不需要的数据，然后我们代理的父原型的功能。

So we overwrite save function on the model instance, but we just filter out the data we don't need, and then we proxy that to the parent prototype function.

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