MongoDB-查询两个集合

Question

I am trying to get hold of MongoDB and have been experimenting with some basic queries on a single collection but I would like to use two collections in my queries by linking them somehow. I have the below document with information of a movie,

Movies:

{
      id:  1, 
      title: "abc", 
      release_data: "xxxx",
      IMDBURL: "qwe", 
      genre: ["xx","yy's","zz"]
}

and the below type of document in another collection which has information about a user and an embedded document with the movies he has rated as well as the rating itself.

Users:

{
      id:  1, 
      age: xx, 
      gender: "Y", 
      occupation: "abc", 
      zip_code: "asd", 
      movies:[
      { movie: 1, rating: 5 } , 
      { movie: 2, rating: 3 }, 
      { movie: 3, rating: 4 }, 
      { movie: 4, rating: 3 }
      ]
}

How would I make a query that returns the title of a movie that has been rated a 5 at least one time? Thank you.

Answer 1

MongoDB: No JOINS, no transactions

Funny enough, I have never needed either of them. As with your example, you basically have to ask yourself what you need to have answered and model your data accordingly .

• Which movies have been rated at least five times?
• For those movies, what are the names?

Given your data model, you don't even get away with ordinary queries: you need an aggregation. With the following dataset:

{ "_id" : ObjectId("56c8e58ee99e5c4e87ec3a4e"), "age" : 22, "gender" : "Y", "occupation" : "abc", "zip_code" : "asd", "movies" : [ { "movie" : 1, "rating" : 5 }, { "movie" : 2, "rating" : 3 }, { "movie" : 3, "rating" : 4 }, { "movie" : 4, "rating" : 3 } ] }
{ "_id" : ObjectId("56c8e598e99e5c4e87ec3a4f"), "age" : 22, "gender" : "Y", "occupation" : "abc", "zip_code" : "asd", "movies" : [ { "movie" : 1, "rating" : 5 }, { "movie" : 2, "rating" : 3 }, { "movie" : 3, "rating" : 4 } ] }
{ "_id" : ObjectId("56c8e599e99e5c4e87ec3a50"), "age" : 22, "gender" : "Y", "occupation" : "abc", "zip_code" : "asd", "movies" : [ { "movie" : 1, "rating" : 5 }, { "movie" : 2, "rating" : 3 }, { "movie" : 3, "rating" : 4 } ] }
{ "_id" : ObjectId("56c8e59ae99e5c4e87ec3a51"), "age" : 22, "gender" : "Y", "occupation" : "abc", "zip_code" : "asd", "movies" : [ { "movie" : 1, "rating" : 5 }, { "movie" : 2, "rating" : 3 }, { "movie" : 3, "rating" : 4 } ] }
{ "_id" : ObjectId("56c8e59be99e5c4e87ec3a52"), "age" : 22, "gender" : "Y", "occupation" : "abc", "zip_code" : "asd", "movies" : [ { "movie" : 1, "rating" : 5 }, { "movie" : 2, "rating" : 3 }, { "movie" : 3, "rating" : 4 } ] }

You can find the movies matching your criteria with the following aggregation

db.movies.aggregate([
  { $unwind:"$movies"},
  { $group:{ _id:"$movies.movie", ratings:{ $sum:1 }}},
  { $match:{ ratings:{ $gte:5}}},
  { $project:{ _id:1 }}
])

Which will return documents looking like this

{ "_id" : 3 }
{ "_id" : 2 }
{ "_id" : 1 }

matching the sample data above. Now, with those, you can look up the movie names in the according collection.

The aggregation, dissected

db.movies.aggregate([

• { $unwind:"$movies"}, : Array contents can nor be inspected directly in a way helpful to us here, so we need to deconstruct the array. For each array item, the parent document is basically copied.
• { $group:{ _id:"$movies.movie", ratings:{ $sum:1 }}}, : We increment the counter ratings , each time a particular movie is found. So each time a movie rating is found for a particular movie, the counter is incremented.
• { $match:{ ratings:{ $gte:5}}}, : we only want documents which have been rated at least 5 times
• { $project:{ _id:1 }} : Last but not least, since we only want the IDs of the movies, we remove everything else.

])

To see what the individual stage does, simply remove the stages following it.