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Javascript 从Spine.js模型的实例获取模型名称_Javascript_Spine.js - Fatal编程技术网

Javascript 从Spine.js模型的实例获取模型名称

Javascript 从Spine.js模型的实例获取模型名称,javascript,spine.js,Javascript,Spine.js,是否可以从spine js模型的实例中获取模型名称?例如,假设我有以下对象: var Client = Spine.Model.sub(); Client.configure('Client', 'id', 'name'); 如果我将对象的实例传递给一个方法,是否可以获取模型名 var client = new Client([id: '0', name: 'Anne']); derp(client); 比如: function derp(c){ c.class() => 'Cli

是否可以从spine js模型的实例中获取模型名称?例如,假设我有以下对象:

var Client = Spine.Model.sub();
Client.configure('Client', 'id', 'name');
如果我将对象的实例传递给一个方法,是否可以获取模型名

var client = new Client([id: '0', name: 'Anne']);
derp(client);
比如:

function derp(c){
  c.class() => 'Client'
}
你可以试试

Object.getPrototypeOf(client).constructor.className


\uuuuu proto\uuuuu
已被弃用,今后将从javascript中删除,但可在旧浏览器中使用
Object.getPrototypeOf
应该在现代浏览器中工作。

是的,它是name属性:Client.name

我想从类的实例中获取类名。
client.__proto__.constructor.className