$project stage for shaping output in MongoDB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

What is the output of this $project stage?

Given the collection students with documents like { name: "Alice", scores: { math: 90, english: 85 }, age: 20 }, what will be the output of this aggregation pipeline stage?

{ $project: { name: 1, mathScore: "$scores.math", _id: 0 } }

A [{ "name": "Alice", "scores": { "math": 90, "english": 85 } }]

B [{ "name": "Alice", "mathScore": 90 }]

C [{ "name": "Alice", "mathScore": { "$scores.math": 90 } }]

D [{ "name": "Alice", "mathScore": "$scores.math" }]

💡 Hint

Remember that $project reshapes documents and you can rename fields using expressions.

The $project stage includes the name field and creates a new field mathScore by extracting scores.math. The _id is excluded. So the output shows only name and mathScore with the math score value.

📝 Syntax

intermediate

Which $project stage is syntactically correct?

Choose the valid $project stage syntax to include title and rename author.name to authorName.

A { $project: { title: 1, authorName: "$author.name" } }

B { $project: { title: 1, authorName: author.name } }

C { $project: { title: 1, authorName: "author.name" } }

D { $project: { title: 1, authorName: $author.name } }

💡 Hint

Field paths must be strings starting with $ inside $project.

In $project, to rename a field using a nested field, you must use a string with a $ prefix like "$author.name". Option A correctly uses this syntax. Other options either miss quotes or the $ sign.

🧠 Conceptual

advanced

What does excluding _id in $project do?

In MongoDB aggregation, what is the effect of setting _id: 0 in a $project stage?

A It causes a syntax error because <code>_id</code> cannot be excluded.

B It renames the <code>_id</code> field to zero.

C It includes the <code>_id</code> field with value zero.

D It removes the <code>_id</code> field from the output documents.

💡 Hint

Think about what _id normally does in MongoDB documents.

By default, _id is included in all documents. Setting _id: 0 in $project excludes it from the output documents.

❓ query_result

advanced

What is the output of this $project with computed field?

Given documents { price: 100, quantity: 3 }, what is the output of this $project stage?

{ $project: { total: { $multiply: ["$price", "$quantity"] }, _id: 0 } }

A [{ "total": 300 }]

B [{ "total": "$price * $quantity" }]

C [{ "total": { "$multiply": [100, 3] } }]

D [{ "price": 100, "quantity": 3, "total": 300 }]

💡 Hint

The $multiply operator calculates the product of the fields.

The $project stage computes a new field total by multiplying price and quantity. The output includes only total because _id is excluded and other fields are not included.

🔧 Debug

expert

Why does this $project stage cause an error?

Consider this $project stage:

{ $project: { name: 1, age: "$age", age: 1 } }

Why will this cause an error or unexpected behavior?

A Because <code>_id</code> is missing and must always be included.

B Because <code>age: "$age"</code> is invalid syntax in $project.

C Because the field <code>age</code> is defined twice, causing a duplicate key error.

D Because <code>name: 1</code> must come after <code>age</code> fields.

💡 Hint

Check the keys in the object carefully for duplicates.

In JSON and BSON, keys must be unique. Defining age twice in the same object causes a duplicate key error or the last one overwrites the first, leading to unexpected results.