有没有办法使用聚合在一个语句中写这个？

Question

使用 movieDetails 集合，计算获得至少 1 个奖项并获得 2 个以上提名的所有电影的最高、最低和平均值 imdb.rating。

数据示例：

db.movieDetails.find().pretty().limit(1)
{
"_id" : ObjectId("5e36f5195333e25b75cbe8cf"),
"title" : "West Side Story",
"year" : 1961,
"rated" : "UNRATED",
"runtime" : 152,
"countries" : [
"USA"
],
"genres" : [
"Crime",
"Drama",
"Musical"
],
"director" : "Jerome Robbins, Robert Wise",
"writers" : [
"Ernest Lehman",
"Arthur Laurents",
"Jerome Robbins"
],
"actors" : [
"Natalie Wood",
"Richard Beymer",
"Russ Tamblyn",
"Rita Moreno"
],
"plot" : "Two youngsters from rival New York City gangs fall in love, but tensions between their respective friends build toward tragedy.",
"poster" : "http://ia.media-imdb.com/images/M/MV5BMTM0NDAxOTI5MF5BMl5BanBnXkFtZTcwNjI4Mjg3NA@@._V1_SX300.jpg",
"imdb" : {
"id" : "tt0055614",
"rating" : 7.6,
"votes" : 67824
},
"awards" : {
"wins" : 18,
"nominations" : 11,
"text" : "Won 10 Oscars. Another 18 wins & 11 nominations."
},
"type" : "movie"
}

这是我所做的。

Highest Rating :

db.movieDetails.find({'awards.wins':{$gte:1},'awards.nominations':{$gte:2}}).sort({'imdb.rating':+1}).limit(1)
Lowest Rating :

db.movieDetails.find({'awards.wins':{$gte:1},'awards.nominations':{$gte:2}}).sort({'imdb.rating':-1}).limit(1)
Avg Rating :

db.movieDetails.find({'awards.wins':{$gte:1},'awards.nominations':{$gte:2}}).aggregate([ { "$group": { "avgOfRating":{$avg": imdb.rating } }]);

我也试过聚合但收到错误..这是我需要帮助的地方。

db.movieDetails.aggregate([{
$project:
    {
    _id:0,
    title:1,
    "imdb.rating":1,
    "awards.wins":1,
    "awards.nominations":1
    },
    $match{"$awards.nominations": {"$gte": 2}, "awards.wins": {"$gte": 1}  }
    }]);

Answer 1

可能是这个：

db.movieDetails.aggregate([
   { $match: { 'awards.wins': { $gte: 1 }, 'awards.nominations': { $gt: 2 } } },
   {
      $group: {
         _id: null,
         highest: { $max: "$imdb.rating" },
         lowest: { $min: "$imdb.rating" },
         average: { $avg: "$imdb.rating" },
      }
   }
])

“获得超过 2 项提名”将是 > 2，而不是 >= 2

有没有办法使用聚合在一个语句中写这个？

Is there a way to write this in one statement using aggregation?

aggregate-functions

mongodb

mongodb-query