Question

之前我问过question 。问题

    {
    "_id" : ObjectId("5539d45ee3cd0e48e99c3fa6"),
    "userId" : 1,
    "movieId" : 6,
    "rating" : 2.0000000000000000,
    "timestamp" : 9.80731e+008
}

    {
    "_id" : ObjectId("5539d45ee3cd0e48e99c1fa7"),
    "userId" : 1,
    "movieId" : 22,
    "rating" : 3.0000000000000000,
    "timestamp" : 9.80731e+008
},

{
    "_id" : ObjectId("5539d45ee3cd0e48e99c1fa8"),
    "userId" : 1,
    "movieId" : 32,
    "rating" : 2.0000000000000000,
    "timestamp" : 9.80732e+008
},


{
    "_id" : ObjectId("5539d45ee3cd0e48e99c1fa9"),
    "userId" : 2,
    "movieId" : 32,
    "rating" : 4.0000000000000000,
    "timestamp" : 9.80732e+008
},

{
    "_id" : ObjectId("5539d45ee3cd0e48e99c1fa3"),
    "userId" : 2,
    "movieId" : 6,
    "rating" : 5.0000000000000000,
    "timestamp" : 9.80731e+008
}

然后需要为给定的两个用户（如userId：1和userId：2）获取公共（交叉）项，如[6,32]。

但是现在我需要得到它们的评分，如 [{＆＃34; movieId＆＃34;：6，＆＃34; user1_rating＆＃34; ：2，＆＃34; user2_rating＆＃34; ：4}，{＆＃34; movieId＆＃34;：32，＆＃34; user1_rating＆＃34; ：2，＆＃34; user2_rating＆＃34; ：5}]

我怎么能得到它？我尝试用

    db.collection.aggregate([
  {$match: {"$or":[{"userId":2},{"userId":1}]}},
  {$group: {_id: "$movieId", users: {$push: {"userId":"$userId","rating":"$rating"}}}},
  {$project: { movieId: "$_id", _id: 0,rating:"$users.rating", allUsersIncluded: { $setIsSubset: [ [1,2], "$users.userId"]}}},
  {$match: { allUsersIncluded: true }},
  {$group: { _id: null, movies: {$push: {"movie":"$movieId","Rating":"$rating"}}}}
])

但我得到[{＆＃34;电影＆＃34;：6,0：2,1：4}，{＆＃34;电影＆＃34;：32,0：2,1：5}] < / p>

Answer 1

最后我实现了我的目标。答案是

    db.collection.aggregate([
  {$match: {"$or":[{"userId":2},{"userId":1}]}},
  {$group: {_id: "$movieId", users: {$addToSet: {"userId":"$userId","rating":"$rating"}}}},
  {$project: { movieId: "$_id", _id: 0,user:"$users", allUsersIncluded: { $setIsSubset: [ [1,2], "$users.userId"]}}},
  {$match: { allUsersIncluded: true }},
  {$group: { _id: null, movies: {$addToSet: {"movie":"$movieId","user":"$user"}}}}
])

Mongodb聚合 - 首先创建项目列表并获得具有评级详细信息的项目的交叉

1 个答案: