问jQuery - $.extend未返回预期函数

EN

api.dom.compose = function(element) {
    // stuff
}

api.dom.email = function(element) {
    this.id = element;
    var message_class_id = 'm' + this.id;
    this.id_element = $('div.ii.gt div.a3s.aXjCH.' + message_class_id);
    element = this.id_element.closest('div.adn');

    this.$el = element;
    return this;
};

$.extend(api.dom.email.prototype, {
    body: function(body) {
        var el = this.dom('body');
        if (body) {
            el.html(body);
        }
        return el.html();
    },

    from: function(email, name) {
        var el = this.dom('from');
        if (email) {
            el.attr('email',email);
        }
        if (name) {
            el.attr('name',name);
            el.html(name);
        }
        return {
            email: el.attr('email'),
            name: el.attr('name'),
            el: el
        };
    },

    // more extended functions

});

// more functions on the api.dom object

return api;

var email = provider.dom.email(mId);
console.log(email);