$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.

Practice

1. What does the $project stage do in a MongoDB aggregation pipeline?

easy

A. It filters documents based on conditions.

B. It shapes the output by selecting, renaming, or creating fields.

C. It sorts the documents in ascending or descending order.

D. It groups documents by a specified key.

Solution

1. Step 1: Understand the purpose of $project

   The $project stage is used to control which fields appear in the output documents and how they are shaped.

2. Step 2: Compare with other stages

   Filtering is done by $match, sorting by $sort, and grouping by $group. So $project is for shaping output fields.

3. Final Answer:

   It shapes the output by selecting, renaming, or creating fields. -> Option B

4. Quick Check:

   $project shapes output = A [OK]

Hint: Remember: $project controls output fields only [OK]

Common Mistakes:

• Confusing $project with $match for filtering
• Thinking $project sorts documents
• Mixing $project with $group for aggregation

2. Which of the following is the correct syntax to include only the name and age fields in the output using $project?

easy

A. { $project: { name: 1, age: 1 } }

B. { $project: { name: 0, age: 0 } }

C. { $project: { name: true, age: true } }

D. { $project: { name: 'include', age: 'include' } }

Solution

1. Step 1: Recall $project syntax for inclusion

   To include fields, set them to 1. To exclude, set to 0. Boolean or strings are invalid.

2. Step 2: Check each option

   { $project: { name: 1, age: 1 } } uses 1 for inclusion correctly. { $project: { name: 0, age: 0 } } excludes fields. { $project: { name: true, age: true } } uses true which is invalid. { $project: { name: 'include', age: 'include' } } uses strings which is invalid.

3. Final Answer:

   { $project: { name: 1, age: 1 } } -> Option A

4. Quick Check:

   Include fields = 1 in $project [OK]

Hint: Use 1 to include fields, 0 to exclude in $project [OK]

Common Mistakes:

• Using true/false instead of 1/0
• Mixing inclusion and exclusion in same $project
• Using strings instead of numbers for field selection

3. Given the collection documents:

{ "name": "Alice", "age": 25, "city": "NY" }

What will be the output of this aggregation?

[{ $project: { fullName: "$name", ageInMonths: { $multiply: ["$age", 12] }, _id: 0 } }]

medium

A. [{ "fullName": "Alice", "ageInMonths": 25 }]

B. [{ "name": "Alice", "ageInMonths": 300 }]

C. [{ "fullName": "Alice", "age": 25 }]

D. [{ "fullName": "Alice", "ageInMonths": 300 }]

Solution

1. Step 1: Understand field renaming and calculation

   The fullName field is created by copying name. The ageInMonths is calculated by multiplying age by 12 (25*12=300). The _id is excluded.

2. Step 2: Check output fields

   Output has fullName with "Alice" and ageInMonths with 300. No name or age fields remain.

3. Final Answer:

   [{ "fullName": "Alice", "ageInMonths": 300 }] -> Option D

4. Quick Check:

   Rename + multiply age = output [OK]

Hint: Use $multiply to calculate new fields in $project [OK]

Common Mistakes:

• Expecting original fields to remain without including them
• Confusing field names after renaming
• Forgetting to exclude _id if not wanted

4. Identify the error in this $project stage:

{ $project: { fullName: "$name", ageInMonths: { $multiply: ["age", 12] }, _id: 0 } }

medium

A. The _id field cannot be excluded.

B. Cannot use $multiply inside $project.

C. Field name "age" should be prefixed with $ to reference the field.

D. The field fullName cannot be renamed.

Solution

1. Step 1: Check field references in expressions

   In MongoDB aggregation, to reference a field inside expressions, prefix with $. Here, "age" is missing $.

2. Step 2: Validate other parts

   $multiply is valid in $project. Excluding _id is allowed. Renaming fields is allowed.

3. Final Answer:

   Field name "age" should be prefixed with $ to reference the field. -> Option C

4. Quick Check:

   Field references need $ prefix [OK]

Hint: Always prefix field names with $ inside expressions [OK]

Common Mistakes:

• Omitting $ before field names in expressions
• Thinking $multiply is invalid in $project
• Believing _id cannot be excluded

5. You have documents with fields firstName, lastName, and birthYear. Write a $project stage that outputs fullName as concatenation of first and last names separated by a space, and age calculated as 2024 minus birthYear. Which is the correct stage?

hard

A. { $project: { fullName: { $concat: ["$firstName", " ", "$lastName"] }, age: { $subtract: [2024, "$birthYear"] }, _id: 0 } }

B. { $project: { fullName: "$firstName $lastName", age: 2024 - "$birthYear", _id: 0 } }

C. { $project: { fullName: { $concat: ["firstName", " ", "lastName"] }, age: { $subtract: [2024, "birthYear"] }, _id: 0 } }

D. { $project: { fullName: { $concat: ["$firstName", "$lastName"] }, age: { $subtract: ["birthYear", 2024] }, _id: 0 } }

Solution

1. Step 1: Build fullName using $concat with spaces

   Use $concat with "$firstName", a space string, and "$lastName" to join names with a space.

2. Step 2: Calculate age using $subtract

   Subtract birthYear field from 2024 using $subtract. Field references need $.

3. Step 3: Check each option

   { $project: { fullName: { $concat: ["$firstName", " ", "$lastName"] }, age: { $subtract: [2024, "$birthYear"] }, _id: 0 } } correctly uses $concat with space and $subtract with field references. { $project: { fullName: "$firstName $lastName", age: 2024 - "$birthYear", _id: 0 } } uses invalid string concatenation and arithmetic. { $project: { fullName: { $concat: ["firstName", " ", "lastName"] }, age: { $subtract: [2024, "birthYear"] }, _id: 0 } } misses $ on fields. { $project: { fullName: { $concat: ["$firstName", "$lastName"] }, age: { $subtract: ["birthYear", 2024] }, _id: 0 } } misses space and reverses subtraction.

4. Final Answer:

   { $project: { fullName: { $concat: ["$firstName", " ", "$lastName"] }, age: { $subtract: [2024, "$birthYear"] }, _id: 0 } } -> Option A

5. Quick Check:

   Use $concat with space and $subtract with $field [OK]

Hint: Use $concat with spaces and $subtract with $field for calculations [OK]

Common Mistakes:

• Forgetting $ before field names
• Concatenating fields without spaces
• Subtracting in wrong order
• Using string expressions instead of operators