如何在 MongoDB 中仅获取数组元素而不是整个对象作为输出？答案

【问题标题】：How can I get only the array element as output instead of whole object in MongoDB?如何在 MongoDB 中仅获取数组元素而不是整个对象作为输出？
【发布时间】：2021-12-11 04:28:49
【问题描述】：

以下是我的代码，用于显示作为餐厅集合对象一部分的评论数组数据：

async get(reviewId) {
    const restaurantsCollection = await restaurants();
    reviewId = ObjectId(reviewId)
  
    const r = await restaurantsCollection.findOne(
      { reviews: { $elemMatch: { _id : reviewId } } },
      {"projection" : { "reviews.$": true }}
    )
  
    return r
  }

我的对象看起来像：

{
  _id: '6176e58679a981181d94dfaf',
  name: 'The Blue Hotel',
  location: 'Noon city, New York',
  phoneNumber: '122-536-7890',
  website: 'http://www.bluehotel.com',
  priceRange: '$$$',
  cuisines: [ 'Mexican', 'Italian' ],
  overallRating: 0,
  serviceOptions: { dineIn: true, takeOut: true, delivery: true },
  reviews: []
}

我的输出如下：

{
    "_id": "6174cfb953edbe9dc5054f99", // restaurant Id
    "reviews": [
        {
            "_id": "6176df77d4639898b0c155f0", // review Id
            "title": "This place was great!",
            "reviewer": "scaredycat",
            "rating": 5,
            "dateOfReview": "10/13/2021",
            "review": "This place was great! the staff is top notch and the food was delicious!  They really know how to treat their customers"
        }
    ]
}

我想要的输出：

{
    "_id": "6176df77d4639898b0c155f0",
    "title": "This place was great!",
    "reviewer": "scaredycat",
    "rating": 5,
    "dateOfReview": "10/13/2021",
    "review": "This place was great! the staff is top notch and the food was delicious!  They really know how to treat their customers"
}

如何在不获取餐厅 ID 或整个对象的情况下仅获取评论的输出？

【问题讨论】：

标签： node.js arrays mongodb objectid

【解决方案1】：

因此查询运算符find 和findOne 不允许“高级”数据重组。

所以你有两种选择：

更常见的方法是在代码中执行此操作，通常人们要么使用某些东西 mongoose post trigger，要么使用某种“共享”函数来处理所有这些转换，这就是避免代码重复的方法。
使用聚合框架，如下所示：

const r = await restaurantsCollection.aggregate([
    {
        $match: { reviews: { $elemMatch: { _id : reviewId } } },
    },
    {
        $replaceRoot: {
            newRoot: {
                $arrayElemAt: [
                    {
                        $filter: {
                            input: "$reviews",
                            as: "review",
                            cond: {$eq: ["$$review._id", reviewId]}
                        }
                    },
                    0
                ]
            }
        }
    }
])
return r[0]

【讨论】：