MongoDB：统计集合中指定对象的总数量

Got it, let's break down how to calculate the total number of times obj1, obj2, and obj3 appear across your collection. Since a single document can have multiple of these objects (like your first document with both obj1 and obj2), we'll use MongoDB's aggregation framework to tally them up accurately.

Aggregation Query #

db.yourCollectionName.aggregate([
  // Step 1: Count how many target objects exist in each document
  {
    $project: {
      documentCount: {
        $sum: [
          { $cond: [{ $exists: ["$obj1", true] }, 1, 0] },
          { $cond: [{ $exists: ["$obj2", true] }, 1, 0] },
          { $cond: [{ $exists: ["$obj3", true] }, 1, 0] }
        ]
      }
    }
  },
  // Step 2: Sum all document-level counts to get the total
  {
    $group: {
      _id: null,
      totalOccurrences: { $sum: "$documentCount" }
    }
  },
  // Optional: Clean up the output to hide the default _id field
  {
    $project: {
      _id: 0
    }
  }
])

How This Works #

• $project Stage: For every document, we use $exists to check if each target object is present. If it is, we add 1 to the documentCount; if not, we add 0. We then sum those values to get how many of the target objects are in that single document.
• $group Stage: We group all documents into one single group (using _id: null) and sum all the documentCount values together to get the grand total of occurrences.
• Final $project Stage: This just removes the auto-generated _id field from the output to make it cleaner.

Example Output #

For your sample documents, this query will return exactly what you're expecting:

{ "totalOccurrences" : 5 }

Quick Notes #

• Don't forget to replace yourCollectionName with the actual name of your MongoDB collection.
• This method works even if the objects have nested fields — $exists checks for the presence of the field itself, regardless of what's inside it.

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