How to display a specific field in array using $project in MongoDB and ignore other fields?

To display a specific field from an array in MongoDB and ignore other fields, use the $project operator with $unwind. Set unwanted fields to 0 to exclude them from the output.

Syntax

db.collection.aggregate([
    { $unwind: "$arrayField" },
    { $match: { "arrayField.field": "value" } },
    { $project: { "_id": 0, "arrayField.unwantedField": 0 } }
]);

Sample Data

db.demo731.insertOne({
    "ProductInformation": [
        { "ProductId": "Product-1", "ProductPrice": 80 },
        { "ProductId": "Product-2", "ProductPrice": 45 },
        { "ProductId": "Product-3", "ProductPrice": 50 }
    ]
});

{
    "acknowledged": true,
    "insertedId": ObjectId("5eac5efd56e85a39df5f6341")
}

Display All Documents

db.demo731.find();

{
    "_id": ObjectId("5eac5efd56e85a39df5f6341"),
    "ProductInformation": [
        { "ProductId": "Product-1", "ProductPrice": 80 },
        { "ProductId": "Product-2", "ProductPrice": 45 },
        { "ProductId": "Product-3", "ProductPrice": 50 }
    ]
}

Example: Display ProductId and Hide ProductPrice

Query to display only ProductId for products with price 80 ?

db.demo731.aggregate([
    { $unwind: "$ProductInformation" },
    { $match: { "ProductInformation.ProductPrice": 80 } },
    { $project: { "_id": 0, "ProductInformation.ProductPrice": 0 } }
]);

{ "ProductInformation": { "ProductId": "Product-1" } }

How It Works

• $unwind separates each array element into individual documents
• $match filters documents based on array field criteria
• $project controls field visibility: 0 excludes, 1 includes

Conclusion

Use $project with $unwind to selectively display array fields in MongoDB. Set unwanted fields to 0 to exclude them, keeping only the required data in your aggregation results.