$lookup for joining collections in MongoDB - Mini Project: Build & Apply

$lookup for joining collections

📖 Scenario: You work at a small online bookstore. You have two collections: books and authors. Each book has an author_id that matches the _id of an author. You want to combine information from both collections to see each book with its author's details.

🎯 Goal: Build a MongoDB aggregation query using $lookup to join the books collection with the authors collection, so each book document includes its author's information.

📋 What You'll Learn

Create a books collection with three books, each having _id, title, and author_id fields.

Create an authors collection with three authors, each having _id and name fields.

Write an aggregation query on books using $lookup to join with authors on author_id and _id.

Add a $project stage to show only title and the joined author name.

💡 Why This Matters

🌍 Real World

Joining collections is common when you have related data in different places, like customers and orders or products and suppliers.

💼 Career

Understanding $lookup is essential for MongoDB developers to combine data efficiently without duplicating it.

1

Create the books collection

Create a books collection with these exact documents: { _id: 1, title: "The Great Gatsby", author_id: 101 }, { _id: 2, title: "1984", author_id: 102 }, and { _id: 3, title: "To Kill a Mockingbird", author_id: 103 }.

MongoDB

// Insert the three book documents into the books collection
// Your code here

Hint

Use db.books.insertMany() with an array of objects for the three books.

2

Create the authors collection

Create an authors collection with these exact documents: { _id: 101, name: "F. Scott Fitzgerald" }, { _id: 102, name: "George Orwell" }, and { _id: 103, name: "Harper Lee" }.

MongoDB

db.books.insertMany([
  { _id: 1, title: "The Great Gatsby", author_id: 101 },
  { _id: 2, title: "1984", author_id: 102 },
  { _id: 3, title: "To Kill a Mockingbird", author_id: 103 }
])
// Insert the three author documents into the authors collection
// Your code here

Hint

Use db.authors.insertMany() with an array of author objects.

3

Write the aggregation query with $lookup

Write an aggregation query on books using db.books.aggregate() with a $lookup stage that joins authors on author_id from books matching _id from authors. Use as: "author_info" to store the joined data.

MongoDB

db.books.insertMany([
  { _id: 1, title: "The Great Gatsby", author_id: 101 },
  { _id: 2, title: "1984", author_id: 102 },
  { _id: 3, title: "To Kill a Mockingbird", author_id: 103 }
])
db.authors.insertMany([
  { _id: 101, name: "F. Scott Fitzgerald" },
  { _id: 102, name: "George Orwell" },
  { _id: 103, name: "Harper Lee" }
])

// Write the aggregation query with $lookup below
// Your code here

Hint

Use $lookup inside db.books.aggregate() with the correct fields and collection names.

4

Add $project to show book title and author name

Add a $project stage after $lookup in the aggregation pipeline to show only the title field and the first element's name from author_info as author.

MongoDB

db.books.insertMany([
  { _id: 1, title: "The Great Gatsby", author_id: 101 },
  { _id: 2, title: "1984", author_id: 102 },
  { _id: 3, title: "To Kill a Mockingbird", author_id: 103 }
])
db.authors.insertMany([
  { _id: 101, name: "F. Scott Fitzgerald" },
  { _id: 102, name: "George Orwell" },
  { _id: 103, name: "Harper Lee" }
])
db.books.aggregate([
  {
    $lookup: {
      from: "authors",
      localField: "author_id",
      foreignField: "_id",
      as: "author_info"
    }
  },
  // Add $project stage here
  // Your code here
])

Hint

Use $project to keep title and extract the first author's name from author_info array.

Practice

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

easy

A. It deletes documents from a collection.

B. It joins documents from two collections based on a matching field.

C. It sorts documents in ascending order.

D. It updates documents with new values.

Solution

1. Step 1: Understand the purpose of $lookup

   $lookup is used to combine documents from two collections by matching fields.

2. Step 2: Compare with other options

   Deleting, sorting, or updating documents are different operations and not related to $lookup.

3. Final Answer:

   It joins documents from two collections based on a matching field. -> Option B

4. Quick Check:

   $lookup joins collections = A [OK]

Hint: Remember: $lookup is like a join in SQL [OK]

Common Mistakes:

• Confusing $lookup with update or delete operations
• Thinking $lookup sorts documents
• Assuming $lookup modifies documents instead of joining

2. Which of the following is the correct syntax for a basic $lookup stage in MongoDB?

easy

A. { $lookup: { from: 'other', localField: 'id', foreignField: 'userId', joinAs: 'joinedData' } }

B. { $lookup: { collection: 'other', localKey: 'id', foreignKey: 'userId', output: 'joinedData' } }

C. { $lookup: { from: 'other', localField: 'id', foreignField: 'userId', as: 'joinedData' } }

D. { $lookup: { from: 'other', localField: 'id', foreignField: 'userId' } }

Solution

1. Step 1: Identify correct field names in $lookup

   The correct keys are from, localField, foreignField, and as.

2. Step 2: Check each option

   { $lookup: { from: 'other', localField: 'id', foreignField: 'userId', as: 'joinedData' } } uses all correct keys and syntax. Options A, B, and C have incorrect or missing keys.

3. Final Answer:

   { $lookup: { from: 'other', localField: 'id', foreignField: 'userId', as: 'joinedData' } } -> Option C

4. Quick Check:

   Correct $lookup keys = D [OK]

Hint: Use keys: from, localField, foreignField, as [OK]

Common Mistakes:

• Using wrong key names like collection or joinAs
• Omitting the 'as' field for output array
• Missing foreignField or localField keys

3. Given two collections: orders with documents { _id: 1, productId: 101 } and products with documents { _id: 101, name: 'Pen' }, what will be the result of this aggregation on orders?

{ $lookup: { from: 'products', localField: 'productId', foreignField: '_id', as: 'productInfo' } }

medium

A. [{ _id: 1, productId: 101, productInfo: [{ _id: 101, name: 'Pen' }] }]

B. [{ _id: 1, productId: 101, productInfo: [] }]

C. [{ _id: 1, productId: 101 }]

D. SyntaxError

Solution

1. Step 1: Match localField and foreignField values

   The productId in orders is 101, which matches _id 101 in products.

2. Step 2: Understand $lookup output

   The matching product document is added as an array in productInfo.

3. Final Answer:

   [{ _id: 1, productId: 101, productInfo: [{ _id: 101, name: 'Pen' }] }] -> Option A

4. Quick Check:

   Matching product added in array = B [OK]

Hint: Matched docs appear in array named by 'as' [OK]

Common Mistakes:

• Expecting productInfo to be a single object, not an array
• Thinking no match means error instead of empty array
• Ignoring the 'as' field name in output

4. What is wrong with this $lookup stage?

{ $lookup: { from: 'users', localField: 'userId', foreignField: 'id', as: 'userDetails' } }

Assuming the users collection uses _id as the user ID field.

medium

A. The 'as' field cannot be 'userDetails'.

B. The localField should be 'id' instead of 'userId'.

C. The 'from' field should be 'user' not 'users'.

D. The foreignField should be '_id' instead of 'id'.

Solution

1. Step 1: Check foreignField correctness

   The users collection uses _id as the ID field, so foreignField must be '_id'.

2. Step 2: Verify other fields

   localField and as are valid; collection name 'users' is correct.

3. Final Answer:

   The foreignField should be '_id' instead of 'id'. -> Option D

4. Quick Check:

   foreignField must match actual field = C [OK]

Hint: Match foreignField to actual field name in other collection [OK]

Common Mistakes:

• Using wrong foreignField name
• Changing 'as' field unnecessarily
• Mistaking collection name pluralization

5. You have two collections: students with { _id: 1, name: 'Alice' } and grades with multiple documents per student like { studentId: 1, score: 85 }. How do you write a $lookup to get each student with all their grades in an array called allGrades?

hard

A. { $lookup: { from: 'grades', localField: '_id', foreignField: 'studentId', as: 'allGrades' } }

B. { $lookup: { from: 'grades', localField: 'studentId', foreignField: '_id', as: 'allGrades' } }

C. { $lookup: { from: 'grades', localField: 'name', foreignField: 'studentId', as: 'allGrades' } }

D. { $lookup: { from: 'grades', localField: '_id', foreignField: 'score', as: 'allGrades' } }

Solution

1. Step 1: Identify matching fields between collections

   students._id matches grades.studentId, so these are the fields to join on.

2. Step 2: Write $lookup with correct fields and output array name

   Use from: 'grades', localField: '_id', foreignField: 'studentId', and as: 'allGrades'.

3. Final Answer:

   { $lookup: { from: 'grades', localField: '_id', foreignField: 'studentId', as: 'allGrades' } } -> Option A

4. Quick Check:

   Match student _id to grades studentId = A [OK]

Hint: Join on matching ID fields, output array named by 'as' [OK]

Common Mistakes:

• Swapping localField and foreignField
• Using wrong fields like name or score for join
• Incorrect output array name