Use MongoDB Aggregate and select only top record (descending)

To select only the top record in descending order using MongoDB aggregate, use the $sort stage to order documents and the $limit stage to restrict the result to one document.

Syntax

db.collection.aggregate([
    { $sort: { fieldName: -1 } },
    { $limit: 1 }
]);

Sample Data

db.demo267.insertMany([
    { id: 100, "Name": "Chris" },
    { id: 100, "Name": "Adam" },
    { id: 100, "Name": "David" },
    { id: 100, "Name": "Bob" }
]);

{
    "acknowledged": true,
    "insertedIds": [
        ObjectId("5e4811951627c0c63e7dbaab"),
        ObjectId("5e48119e1627c0c63e7dbaac"),
        ObjectId("5e4811a51627c0c63e7dbaad"),
        ObjectId("5e4811ab1627c0c63e7dbaae")
    ]
}

View All Documents

db.demo267.find();

{
    "_id": ObjectId("5e4811951627c0c63e7dbaab"),
    "id": 100,
    "Name": "Chris"
}
{
    "_id": ObjectId("5e48119e1627c0c63e7dbaac"),
    "id": 100,
    "Name": "Adam"
}
{
    "_id": ObjectId("5e4811a51627c0c63e7dbaad"),
    "id": 100,
    "Name": "David"
}
{
    "_id": ObjectId("5e4811ab1627c0c63e7dbaae"),
    "id": 100,
    "Name": "Bob"
}

Method 1: Using Aggregate Pipeline

db.demo267.aggregate([
    { $sort: { Name: -1 } },
    { $limit: 1 }
]);

{
    "_id": ObjectId("5e4811a51627c0c63e7dbaad"),
    "id": 100,
    "Name": "David"
}

Method 2: Using find() with sort() and limit()

db.demo267.find().sort({ Name: -1 }).limit(1);

{
    "_id": ObjectId("5e4811a51627c0c63e7dbaad"),
    "id": 100,
    "Name": "David"
}

Key Points

• Use -1 for descending order and 1 for ascending order in $sort.
• The aggregate pipeline approach provides more flexibility for complex operations.
• Both methods return "David" as it comes last alphabetically.

Conclusion

Use $sort with -1 followed by $limit: 1 in an aggregation pipeline to select the top record in descending order. The find().sort().limit() method provides a simpler alternative for basic sorting needs.