$lookup for joining collections in MongoDB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform a $lookup that joins the 'orders' collection with the 'customers' collection on the 'customerId' field.

MongoDB

db.orders.aggregate([{ $lookup: { from: "[1]", localField: "customerId", foreignField: "_id", as: "customerInfo" } }])

Drag options to blanks, or click blank then click option'

Acustomers

Bproducts

Csales

Demployees

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong collection name in the 'from' field.

Confusing 'from' with 'localField' or 'foreignField'.

The from field in $lookup specifies the collection to join with. Here, we join 'orders' with 'customers'.

2fill in blank

medium

Complete the code to join 'orders' with 'products' collection using $lookup on 'productId' and '_id'.

MongoDB

db.orders.aggregate([{ $lookup: { from: "products", localField: "[1]", foreignField: "_id", as: "productDetails" } }])

Drag options to blanks, or click blank then click option'

AcustomerId

BorderId

CproductId

DsupplierId

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'customerId' instead of 'productId' for product join.

Mixing up localField and foreignField.

The localField is the field in the 'orders' collection that matches the '_id' in 'products'. Here, it's 'productId'.

3fill in blank

hard

Fix the error in the $lookup stage to correctly join 'orders' with 'suppliers' on 'supplierId'.

MongoDB

db.orders.aggregate([{ $lookup: { from: "suppliers", localField: "supplierId", foreignField: "[1]", as: "supplierInfo" } }])

Drag options to blanks, or click blank then click option'

AsupplierId

BsupplierID

Cid

D_id

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'supplierId' as foreignField instead of '_id'.

Incorrect casing like 'supplierID'.

The foreignField should be the field in 'suppliers' that matches 'supplierId' in 'orders'. Usually, this is '_id'.

4fill in blank

hard

Fill both blanks to join 'orders' with 'employees' collection on 'employeeId' and project only 'orderId' and 'employeeName'.

MongoDB

db.orders.aggregate([ { $lookup: { from: "[1]", localField: "employeeId", foreignField: "_id", as: "employeeInfo" } }, { $project: { orderId: 1, [2]: "$employeeInfo.name" } } ])

Drag options to blanks, or click blank then click option'

Aemployees

BcustomerName

CemployeeName

Dorders

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong collection name in 'from'.

Projecting with incorrect field names.

The from collection is 'employees'. The projected field name for the employee's name is 'employeeName'.

5fill in blank

hard

Fill all three blanks to join 'orders' with 'categories' on 'categoryId', unwind the joined array, and filter categories with name 'Beverages'.

MongoDB

db.orders.aggregate([ { $lookup: { from: "[1]", localField: "categoryId", foreignField: "_id", as: "categoryInfo" } }, { $unwind: "[2]" }, { $match: { "categoryInfo.[3]": "Beverages" } } ])

Drag options to blanks, or click blank then click option'

Acategories

B$categoryInfo

Cname

Dproducts

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong collection name in 'from'.

Incorrect unwind path without '$'.

Filtering on wrong field name.

The from collection is 'categories'. The $unwind stage uses '$categoryInfo' to flatten the array. The $match filters by the 'name' field inside 'categoryInfo'.

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