如何在 MongoDB 中只獲取數組元素而不是整個對像作為輸出？

游泳巴伊

下面是我的代碼，用於顯示作為餐廳集合對像一部分的評論數組數據：

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

我的對像看起來像：

{
  _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: []
}

我的輸出看起來像：

{
    "_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"
        }
    ]
}

我想要的輸出：

{
    "_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"
}

如何在不獲取餐廳 ID 或整個對象的情況下僅將輸出作為評論？

湯姆·斯拉巴特

所以查詢操作符，find並findOne不允許“高級”的數據重組。

所以你有兩個選擇：

更常見的方法是在代碼中執行此操作，通常人們要么使用某種東西mongoose post trigger要么使用某種“共享”函數來處理所有這些轉換，這就是避免代碼重複的方法。
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]