Mongodb聚合-首先创建项目列表并获取项目与评级详细信息的交集

我以前问过他。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
}

然后需要获得给定两个用户（如userId:1和userId:2）的公共（相交）项，如[6,32]

但是现在我需要用它们中的每一个的评级，比如[{“movieId”：6，“user1_评级”：2，“user2_评级”：4}，{“movieId”：32，“user1_评级”：2，“user2_评级”：5}]

我怎么能得到这个？ 我试着去做

    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"}}}}
])

---

但是我得到了[{“电影”：6,0:2,1:4}，{“电影”：32,0:2,1:5}]

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"}}}}
])