从MongooseModel#discriminator函数创建模型时，如何设置鉴别器关键字的值？

Question

对于Mongoose，我知道我可以使用鉴别器特性，通过一个鉴别器键让不同的模式存在于同一个集合中。

const options = { discriminatorKey: 'kind' };
const Event = mongoose.model('Event, new Schema({
 name: { type: String }
}, options);

const ClickEvent = Event.discriminator('ClickEvent', new Schema({
 url: { type: String }
}, options);

// on another file
const ClickEvent = mongoose.model('ClickEvent');
const clickEvent = new ClickEvent({
 name: 'sir',
 url: 'http://somewhere.com/hello'
});
console.log(clickEvent); // { name: 'sir', url: 'http://somewhere.com/hello', kind: 'ClickEvent' }
// look carefully kind is set to ClickEvent

我希望以某种方式将kind设置为click。我怎么能这样做呢？我希望是这样的：

const ClickEvent = Event.discriminator('ClickEvent', new Schema({
     url: { type: String }
    }, { ...options, discriminatorValue: 'click' });

Answer 1

对于Mongoose 5.2+，您可以将该值作为API Doc中指定的第三个参数进行传递

const ClickEvent = Event.discriminator('ClickEvent', new Schema({
  url: { type: String }
}), "click")

并将kind设置为"click“