JSON aggregation with JSON_AGG in PostgreSQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

What is the output of this JSON_AGG query?

Given a table employees with columns id, name, and department, what is the output of the following query?

SELECT department, JSON_AGG(name ORDER BY name) AS employee_names FROM employees GROUP BY department ORDER BY department;

PostgreSQL

CREATE TABLE employees (id INT, name TEXT, department TEXT);
INSERT INTO employees VALUES
(1, 'Alice', 'HR'),
(2, 'Bob', 'IT'),
(3, 'Charlie', 'HR'),
(4, 'David', 'IT'),
(5, 'Eve', 'Finance');

SELECT department, JSON_AGG(name ORDER BY name) AS employee_names FROM employees GROUP BY department ORDER BY department;

A [{"department":"Finance","employee_names":["Eve"]},{"department":"HR","employee_names":["Alice","Charlie"]},{"department":"IT","employee_names":["Bob","David"]}]

B [{"department":"Finance","employee_names":["Eve"]},{"department":"HR","employee_names":["Charlie","Alice"]},{"department":"IT","employee_names":["David","Bob"]}]

C [{"department":"Finance","employee_names":["Eve"]},{"department":"HR","employee_names":["Alice","Charlie"]},{"department":"IT","employee_names":["David","Bob"]}]

D [{"department":"Finance","employee_names":["Eve"]},{"department":"HR","employee_names":["Charlie","Alice"]},{"department":"IT","employee_names":["Bob","David"]}]

💡 Hint

Remember that JSON_AGG with ORDER BY sorts the aggregated values inside the array.

The query groups employees by department and aggregates their names into JSON arrays sorted alphabetically. So HR has ["Alice", "Charlie"] and IT has ["Bob", "David"].

🧠 Conceptual

intermediate

What does JSON_AGG do in PostgreSQL?

Choose the best description of what the JSON_AGG function does in PostgreSQL.

A Aggregates JSON objects into a single JSON object by merging keys.

B Aggregates multiple rows into a single JSON array.

C Converts a JSON array into multiple rows.

D Converts a JSON string into a JSON object.

💡 Hint

Think about how aggregation functions combine multiple rows.

JSON_AGG collects values from multiple rows and returns them as a JSON array.

📝 Syntax

advanced

Which query correctly aggregates JSON objects per department?

You want to aggregate employee data as JSON objects per department with keys id and name. Which query is correct?

PostgreSQL

Table employees(id INT, name TEXT, department TEXT)

A SELECT department, JSON_AGG(JSON_BUILD_OBJECT('id', id, 'name', name)) AS employees FROM employees GROUP BY department;

B SELECT department, JSON_AGG(ROW(id, name)) AS employees FROM employees GROUP BY department;

C SELECT department, JSON_OBJECT_AGG(id, name) AS employees FROM employees GROUP BY department;

D SELECT department, JSON_AGG({'id': id, 'name': name}) AS employees FROM employees GROUP BY department;

💡 Hint

Remember JSON_BUILD_OBJECT creates JSON objects from key-value pairs.

Option A uses JSON_BUILD_OBJECT to create JSON objects per row, then aggregates them into a JSON array per department. Option A aggregates SQL ROW types, which is not JSON. Option A aggregates key-value pairs but requires unique keys and returns a JSON object, not array. Option A uses invalid syntax for JSON object creation.

🔧 Debug

advanced

Why does this JSON_AGG query raise an error?

Consider this query:

SELECT department, JSON_AGG(name ORDER BY salary) FROM employees GROUP BY department;

Assuming salary is a column in employees, why does this query raise an error?

A Because ORDER BY inside JSON_AGG requires a subquery.

B Because JSON_AGG does not support ORDER BY inside its parentheses.

C Because <code>name</code> is not a valid column in employees.

D Because <code>salary</code> is not in the GROUP BY clause or an aggregate function.

💡 Hint

Check what columns are allowed in SELECT when using GROUP BY.

The error occurs because salary is used inside ORDER BY in JSON_AGG but is neither grouped nor aggregated, violating SQL rules.

❓ optimization

expert

How to optimize JSON_AGG for large datasets?

You have a large sales table with millions of rows. You want to aggregate sales data per region into JSON arrays using JSON_AGG. Which approach is best to optimize performance?

A Use a CROSS JOIN to combine all regions and sales before aggregation.

B Use JSON_AGG without any filtering or indexing to get all data at once.

C Use a subquery to pre-filter rows by date before applying JSON_AGG.

D Use JSON_AGG inside a window function partitioned by region.

💡 Hint

Filtering data early reduces the amount of data JSON_AGG processes.

Pre-filtering with a subquery reduces rows before aggregation, improving performance. CROSS JOIN and window functions here increase data volume or complexity. No filtering causes unnecessary processing.

Practice

1. What does the JSON_AGG function do in PostgreSQL?

easy

A. It collects multiple rows into a JSON array.

B. It converts a JSON array into separate rows.

C. It filters JSON data based on a condition.

D. It updates JSON data in a table.

Solution

1. Step 1: Understand JSON_AGG purpose

   JSON_AGG is an aggregate function that collects multiple rows into a single JSON array.

2. Step 2: Compare options with function behavior

   Only It collects multiple rows into a JSON array. correctly describes this behavior; others describe different functions or actions.

3. Final Answer:

   It collects multiple rows into a JSON array. -> Option A

4. Quick Check:

   JSON_AGG collects rows = It collects multiple rows into a JSON array [OK]

Hint: Remember: AGG means aggregate rows into one [OK]

Common Mistakes:

• Confusing JSON_AGG with JSON functions that extract data
• Thinking JSON_AGG filters rows
• Assuming JSON_AGG updates JSON data

2. Which of the following is the correct syntax to use JSON_AGG in a query?

easy

A. SELECT JSON_AGG(column_name) table_name;

B. SELECT JSON_AGG FROM table_name WHERE column_name;

C. SELECT JSON_AGG(column_name) WHERE table_name;

D. SELECT JSON_AGG(column_name) FROM table_name;

Solution

1. Step 1: Review correct SQL aggregate function syntax

   Aggregate functions like JSON_AGG require parentheses around the column and a FROM clause specifying the table.

2. Step 2: Check each option for syntax correctness

   SELECT JSON_AGG(column_name) FROM table_name; uses correct syntax: SELECT JSON_AGG(column_name) FROM table_name; Others miss FROM or parentheses or have misplaced WHERE.

3. Final Answer:

   SELECT JSON_AGG(column_name) FROM table_name; -> Option D

4. Quick Check:

   Correct aggregate syntax = SELECT JSON_AGG(column_name) FROM table_name; [OK]

Hint: Aggregate functions need parentheses and FROM clause [OK]

Common Mistakes:

• Omitting FROM clause
• Missing parentheses after JSON_AGG
• Misplacing WHERE clause

3. Given the table products with columns category and name, what is the output of this query?

SELECT category, JSON_AGG(name) FROM products GROUP BY category ORDER BY category;

medium

A. [{"category":"Books","json_agg":["Novel","Comics"]}, {"category":"Toys","json_agg":["Car","Doll"]}]

B. [{"category":"Books","json_agg":"Novel,Comics"}, {"category":"Toys","json_agg":"Car,Doll"}]

C. [{"category":"Books","json_agg":["Novel"]}, {"category":"Toys","json_agg":["Car"]}]

D. [{"category":"Books","json_agg":null}, {"category":"Toys","json_agg":null}]

Solution

1. Step 1: Understand GROUP BY with JSON_AGG

   The query groups rows by category and collects all product names per category into a JSON array.

2. Step 2: Analyze expected JSON_AGG output

   JSON_AGG returns an array of names per category, so each category has a JSON array of product names, not a string or null.

3. Final Answer:

   [{"category":"Books","json_agg":["Novel","Comics"]}, {"category":"Toys","json_agg":["Car","Doll"]}] -> Option A

4. Quick Check:

   JSON_AGG groups names into arrays = [{"category":"Books","json_agg":["Novel","Comics"]}, {"category":"Toys","json_agg":["Car","Doll"]}] [OK]

Hint: JSON_AGG returns arrays, not comma-separated strings [OK]

Common Mistakes:

• Expecting JSON_AGG to return strings instead of arrays
• Ignoring GROUP BY effect
• Assuming null results without data

4. Identify the error in this query:

SELECT category, JSON_AGG(name) FROM products GROUP category;

medium

A. Missing FROM clause

B. Incorrect GROUP BY syntax

C. JSON_AGG cannot be used with GROUP BY

D. Missing parentheses in JSON_AGG

Solution

1. Step 1: Check GROUP BY syntax

   The correct syntax is GROUP BY column_name, but the query uses GROUP category without BY.

2. Step 2: Verify other parts of the query

   FROM clause is present, JSON_AGG has parentheses, and it can be used with GROUP BY, so those are correct.

3. Final Answer:

   Incorrect GROUP BY syntax -> Option B

4. Quick Check:

   GROUP BY must include BY keyword = Incorrect GROUP BY syntax [OK]

Hint: GROUP BY always needs the BY keyword [OK]

Common Mistakes:

• Writing GROUP instead of GROUP BY
• Forgetting parentheses in functions
• Assuming JSON_AGG can't be grouped

5. You want to create a JSON object for each department with an array of employee names and their roles. Which query correctly uses JSON_AGG to achieve this?

hard

A. SELECT department, JSON_BUILD_OBJECT('employees', JSON_AGG(name, role)) FROM employees GROUP BY department;

B. SELECT department, JSON_AGG(name, role) FROM employees GROUP BY department;

C. SELECT department, JSON_AGG(JSON_BUILD_OBJECT('name', name, 'role', role)) FROM employees GROUP BY department;

D. SELECT department, JSON_AGG(name || role) FROM employees GROUP BY department;

Solution

1. Step 1: Understand JSON_AGG with JSON_BUILD_OBJECT

   To create JSON objects inside an array, use JSON_BUILD_OBJECT for each row, then aggregate with JSON_AGG.

2. Step 2: Evaluate each option

   SELECT department, JSON_AGG(JSON_BUILD_OBJECT('name', name, 'role', role)) FROM employees GROUP BY department; correctly builds JSON objects per employee and aggregates them by department. Options B and C misuse JSON_AGG arguments, and D concatenates strings incorrectly.

3. Final Answer:

   SELECT department, JSON_AGG(JSON_BUILD_OBJECT('name', name, 'role', role)) FROM employees GROUP BY department; -> Option C

4. Quick Check:

   Use JSON_BUILD_OBJECT inside JSON_AGG for objects [OK]

Hint: Use JSON_BUILD_OBJECT inside JSON_AGG for arrays of objects [OK]

Common Mistakes:

• Passing multiple columns directly to JSON_AGG
• Concatenating strings instead of building JSON objects
• Misusing JSON_BUILD_OBJECT outside aggregation