Question

所以在我的数据库中，我有3个集合，它们看起来像这样：

客户：

customers = [
  {_id: 1, username: "jack", ... },
  {_id: 2, username: "jane", ... }
  ...
]

评论：

reviews = [
  { _id: 1, customerID: 1, message: "my message", ...}
  ...
]

评论：

comments = [
  { _id: 1, reviewID: 1, customerID: 2, message: "my response" ...}
  ...
]

客户可以发表评论，也可以评论其他评论。因此，我想要的是mongodb aggregation查询：

获取评论。
进行审核的客户的数据。
对该评论的评论。
对该评论发表评论的客户的数据。

即

reviews = [
  {
    _id: 1,
    username: "jack",
    message: "my message"
    comments: [
      { _id: 1, username: "jane", message: "my response", ...},
      ...
    ]
    ...
  }
  ...
]

Answer 1

您可以从comments集合开始，到$lookup以customers来获得customer的名称，然后可以$group通过评论查看所有评论，{{3 }}两次（使用reviews和customer）。每当您知道这是一对一的关系时，您可以在$lookup之后使用$unwind。试试：

db.comments.aggregate([
    {
        $lookup: {
            from: "customers",
            localField: "customerID",
            foreignField: "_id",
            as: "customer"
        }
    },
    {
        $unwind: "$customer"
    },
    {
        $project: {
            _id: 1,
            reviewID: 1,
            username: "$customer.username",
            message: 1
        }
    },
    {
        $group: {
            _id: "$reviewID",
            comments: { $push: { _id: "$_id", username: "$username", message: "$message" } }
        }
    },
    {
        $lookup: {
            from: "reviews",
            localField: "_id",
            foreignField: "_id",
            as: "review"
        }
    },
    {
        $unwind: "$review"
    },
    {
        $lookup: {
            from: "customers",
            localField: "review.customerID",
            foreignField: "_id",
            as: "customer"
        }
    },
    {
        $unwind: "$customer"
    },
    {
        $project: {
            _id: 1,
            message: "$review.message",
            username: "$customer.username",
            comments: 1
        }
    }
])

输出：

{ "_id" : 1, "comments" : [ { "_id" : 1, "username" : "jane", "message" : "my response" } ], "message" : "my message", "username" : "jack" }

编辑：如果您想从reviews开始并将其过滤为单个电影，则可以使用$lookup

db.reviews.aggregate([
    {
        $match: {
            movieId: 1,
        }
    },
    {
        $lookup: {
            from: "customers",
            localField: "customerID",
            foreignField: "_id",
            as: "customer"
        }
    },
    {
        $unwind: "$customer"
    },
    {
        $lookup: {
            from: "comments",
            let: { reviewId: "$_id" },
            pipeline: [
                {
                    $match: { $expr: { $eq: [ "$$reviewId", "$reviewID" ] } }
                },
                {
                    $lookup: {
                        from: "customers",
                        localField: "customerID",
                        foreignField: "_id",
                        as: "customer"
                    }
                },
                {
                    $unwind: "$customer"
                },
                {
                    $project: {
                        _id: 1,
                        message: 1,
                        username: "$customer.username"
                    }
                }
            ],
            as: "comments"
        }
    },
    {
        $project: {
            _id: 1,
            message: 1,
            username: "$customer.username",
            comments: 1
        }
    }
])

带来相同的输出

Answer 2

您可以在mongodb 3.6 及更高版本

中使用以下聚合

Reviews.aggregate([
  { "$lookup": {
    "from": Customers.collection.name,
    "let": { "customerID": "$customerID" },
    "pipeline": [
      { "$match": { "$expr": { "$eq": ["$_id", "$$customerID"] } } }
    ],
    "as": "customer"
  }},
  { "$unwind": "$customer" },
  { "$lookup": {
    "from": Comments.collection.name,
    "let": { "reviewID": "$_id" },
    "pipeline": [
      { "$match": { "$expr": { "$eq": ["$reviewID", "$$reviewID"] } } },
      { "$lookup": {
        "from": Customers.collection.name,
        "let": { "customerID": "$customerID" },
        "pipeline": [
          { "$match": { "$expr": { "$eq": ["$_id", "$$customerID"] } } }
        ],
        "as": "customer"
      }},
      { "$unwind": "$customer" },
    ],
    "as": "comments"
  }}
])

$ lookup为子数组中的每个元素

2 个答案: