基于子文档的Mongoose过滤器MongoDB

const mongoose = require('mongoose')
const Review = require('./reviewSchema')

// defining the structore of your document
let dishSchema = mongoose.Schema({
  name : String,
  price :Number,
  imageURL :String,
  reviews : [Review.schema]
})

// convert the schema into a model class which you can use in your code
const Dish = mongoose.model('Dish',dishSchema)

module.exports = Dish


const mongoose = require('mongoose')

let reviewSchema = mongoose.Schema({
  title : String,
  description :String
})

const Review = mongoose.model('Review',reviewSchema)

module.exports = Review 

我的问题是,如果至少有一个评论,我想把所有的菜都拿来。这是我编写的返回空数组的代码。

Dish.find({
  "reviews.length" : { $gt : 0 }
},function(error,dishes){
  console.log(dishes)
})