MongoDB：统计关联集合中被引用文档的出现次数

Got it, let's walk through how to tally up how many times each book from your Books collection gets referenced in the fav_books arrays of the Users collection. MongoDB's aggregation framework is perfect for this job—here's a step-by-step solution:

Solution Code #

db.Users.aggregate([
  // Split each user's fav_books array into individual documents
  { $unwind: "$fav_books" },
  // Group by book ID and count how many times each appears
  {
    $group: {
      _id: "$fav_books",
      count: { $sum: 1 }
    }
  },
  // Transform results into the key-value format you want
  {
    $group: {
      _id: null,
      bookCounts: {
        $push: {
          k: "$_id",
          v: { $toString: "$count" } // Convert count to string to match your example
        }
      }
    }
  },
  {
    $replaceRoot: {
      newRoot: { $arrayToObject: "$bookCounts" }
    }
  }
])

How This Works #

Let's break down each stage of the aggregation pipeline:

• $unwind: Takes every user's fav_books array and splits it into separate documents. For example, a user with fav_books: ["123", "124"] becomes two distinct documents—one for each book ID.
• First $group: Groups all those unwound documents by book ID, then uses $sum: 1 to count how many times each ID shows up across all users.
• Second $group: Collects all the count entries into an array of {k: bookId, v: count} objects. We use $toString here to convert the numeric count to a string, matching your desired output format.
• $replaceRoot: Converts that array of key-value pairs into a single object using $arrayToObject, then sets that object as the final result document.

Expected Output #

Running this aggregation will return exactly the format you're looking for:

{ "123": "3", "124": "2", "125": "2" }

If you ever wanted to include book names from the Books collection too, you could add a $lookup stage after the first $group to join the two collections, then adjust the final transformation to include the name. But for your current requirements, the above code should work flawlessly.

内容的提问来源于stack exchange，提问作者Siraj ul Haq