MongoDB 查询以查找具有共同值的字段和 return 这些字段的所有详细信息

Question

我有一个MongoDBcollection如下：

Collection：电影

[
      {"_id": 0,title: "Movie A",  year: 2010, director: "D1", actors: ["A1","A2"] }
      ,{"_id": 1,title: "Movie B",  year: 2010, director: "D1", actors: ["A1","A5","A6"]}
      ,{"_id": 2,title: "Movie C",  year: 2010, director: "D2", actors: ["A4"]}
      ,{"_id": 3,title: "Movie D",  year: 2012, director: "D3", actors: ["A7"]}
      ,{"_id": 4,title: "Movie E",  year: 2015, director: "D3", actors: ["A7"]}
]

我想return同一导演的所有电影细节，也就是共同导演的记录。

结果应该是这样的：

 {"_id": 0,title: "Movie A",  year: 2010, director: "D1", actors: ["A1","A2"] },
 {"_id": 1,title: "Movie B",  year: 2010, director: "D1", actors: ["A1","A5","A6"]},
 {"_id": 3,title: "Movie D",  year: 2012, director: "D3", actors: ["A7"]},
 {"_id": 4,title: "Movie E",  year: 2015, director: "D3", actors: ["A7"]}

我试过这个：

db.movies.aggregate([{$group:{_id:"$director",count : {$sum : 1}}},{$match:{"count":{$gt:1}}},])

得到了结果：

{ "_id" : "D3", "count" : 2 }
{ "_id" : "D1", "count" : 2 }

但不知道下一步该做什么？

Answer 1

$group by director, 求和得到总数并构造rootDocs 属性
$match过滤count大于1的文档
$unwind解构rootDocs数组
$replaceRoot 将 rootDocs 对象替换为 root

db.collection.aggregate([
  {
    $group: {
      _id: "$director",
      count: { $sum: 1 },
      rootDocs: { $push: "$$ROOT" }
    }
  },
  { $match: { count: { $gt: 1 } } },
  { $unwind: "$rootDocs" },
  { $replaceRoot: { newRoot: "$rootDocs" } }
])

Playground

Answer 2

您可以在我们执行以下管道阶段的地方使用以下聚合：

组：将相同导演的电影放在一起
匹配：寻找共同的导演，即导演应该至少有 2 部电影
展开：因为你需要single/original条记录所以我们需要unwind/ungroup条记录

项目：以预期格式获得结果会将分组数据映射到原始模式。

 [{$group: {
_id: '$director',
commonDirector: {
 $addToSet: '$$ROOT'
},
count:{$sum:1}
}}, {$match: {
count: {$gt:1}
}}, {$unwind: {
path: '$commonDirector',
preserveNullAndEmptyArrays: false
}}, {$project: {
 _id: '$commonDirector._id',
title: '$commonDirector.title',
year:'$commonDirector.year',
director:"$commonDirector.director",
actors:'$commonDirector.actors'
}}, {
$sort: {
 director: 1,
 title:1
   }
}]

然后它会按照您想要的方式生成结果：

 {"_id": 0,title: "Movie A",  year: 2010, director: "D1", actors: ["A1","A2"] },
 {"_id": 1,title: "Movie B",  year: 2010, director: "D1", actors: ["A1","A5","A6"]},
 {"_id": 3,title: "Movie D",  year: 2012, director: "D3", actors: ["A7"]},
 {"_id": 4,title: "Movie E",  year: 2015, director: "D3", actors: ["A7"]}

MongoDB 查询以查找具有共同值的字段和 return 这些字段的所有详细信息

MongoDB query to find fields with common value and return all the details of those fields

mongodb

nosql

aggregation-framework