GROUP BY single column in SQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of GROUP BY on a single column

Given the table Sales with columns Product and Quantity, what is the output of the following query?

SELECT Product, SUM(Quantity) FROM Sales GROUP BY Product;

SQL

CREATE TABLE Sales (Product VARCHAR(20), Quantity INT);
INSERT INTO Sales VALUES ('Apple', 10), ('Banana', 5), ('Apple', 15), ('Banana', 10), ('Cherry', 7);

A [{"Product": "Apple", "SUM(Quantity)": 25}, {"Product": "Banana", "SUM(Quantity)": 15}, {"Product": "Cherry", "SUM(Quantity)": 7}]

B [{"Product": "Apple", "SUM(Quantity)": 10}, {"Product": "Banana", "SUM(Quantity)": 5}, {"Product": "Apple", "SUM(Quantity)": 15}, {"Product": "Banana", "SUM(Quantity)": 10}, {"Product": "Cherry", "SUM(Quantity)": 7}]

C [{"Product": "Apple", "SUM(Quantity)": 25}, {"Product": "Banana", "SUM(Quantity)": 15}]

D [{"Product": "Apple", "SUM(Quantity)": 10}, {"Product": "Banana", "SUM(Quantity)": 15}, {"Product": "Cherry", "SUM(Quantity)": 7}]

💡 Hint

GROUP BY combines rows with the same value in the grouped column and aggregates other columns.

The query groups rows by Product and sums the Quantity for each product. Apple has 10 + 15 = 25, Banana has 5 + 10 = 15, Cherry has 7.

🧠 Conceptual

intermediate

Understanding GROUP BY behavior

What happens if you run this query?

SELECT Product, Quantity FROM Sales GROUP BY Product;

Assuming the same Sales table as before.

A It returns one row per product with a random Quantity value from that product's rows.

B It returns an error because Quantity is not aggregated or in GROUP BY.

C It sums the Quantity for each product automatically.

D It returns all rows without grouping.

💡 Hint

In SQL, columns in SELECT must be aggregated or included in GROUP BY.

SQL requires that all selected columns be either grouped or aggregated. Quantity is neither, so this query causes an error.

📝 Syntax

advanced

Identify the syntax error in GROUP BY query

Which option contains a syntax error when grouping by a single column?

SQL

Table: Orders(OrderID INT, Customer VARCHAR(20), Amount INT)

A SELECT Customer, SUM(Amount) FROM Orders GROUP BY Customer;

B SELECT Customer, SUM(Amount) FROM Orders GROUP BY Customer ORDER BY Customer;

C SELECT Customer, SUM(Amount) FROM Orders GROUP Customer;

D SELECT Customer, SUM(Amount) FROM Orders GROUP BY Customer HAVING SUM(Amount) > 100;

💡 Hint

Check the syntax of the GROUP BY clause.

Option C is missing the keyword BY after GROUP, causing a syntax error.

❓ optimization

advanced

Optimizing GROUP BY query performance

You have a large table Transactions with columns UserID, Amount, and Date. You want to get total amount per user quickly. Which option improves query speed the most?

A Add an index on UserID and run: SELECT UserID, SUM(Amount) FROM Transactions GROUP BY UserID;

B Add an index on Amount and run: SELECT UserID, SUM(Amount) FROM Transactions GROUP BY UserID;

C Run: SELECT UserID, SUM(Amount) FROM Transactions;

D Run: SELECT UserID, SUM(Amount) FROM Transactions GROUP BY UserID ORDER BY UserID;

💡 Hint

Indexes help speed up grouping on the indexed column.

Indexing UserID helps the database quickly group rows by UserID, improving performance.

🔧 Debug

expert

Why does this GROUP BY query return fewer rows than expected?

Given the table Employees with columns Department and Salary, you run:

SELECT Department, COUNT(*) FROM Employees GROUP BY Department;

You notice some departments are missing in the result. What is the most likely reason?

A GROUP BY only shows departments with the highest salary.

B The COUNT(*) function filters out departments with zero employees.

C The query needs a WHERE clause to include all departments.

D Some departments have NULL values in Department column, and GROUP BY excludes NULL groups.

💡 Hint

Think about how NULL values behave in GROUP BY.

GROUP BY treats NULL as a group, but some databases may exclude NULL groups or they appear as NULL. Departments with NULL values might be missing or grouped separately.

Practice

1. What does the GROUP BY clause do in an SQL query?

easy

A. It deletes duplicate rows from the table.

B. It sorts the rows in ascending order.

C. It groups rows that have the same values in a specified column.

D. It filters rows based on a condition.

Solution

1. Step 1: Understand the purpose of GROUP BY

   The GROUP BY clause collects rows with the same value in the specified column into groups.

2. Step 2: Differentiate from other clauses

   Unlike ORDER BY which sorts, or WHERE which filters, GROUP BY organizes data for aggregation.

3. Final Answer:

   It groups rows that have the same values in a specified column. -> Option C

4. Quick Check:

   GROUP BY = grouping rows by column [OK]

Hint: GROUP BY collects rows sharing column values [OK]

Common Mistakes:

• Confusing GROUP BY with ORDER BY
• Thinking GROUP BY filters rows
• Assuming GROUP BY deletes duplicates

2. Which of the following is the correct syntax to group data by the column department?

easy

A. SELECT department, COUNT(*) FROM employees GROUP BY department;

B. SELECT department, COUNT(*) FROM employees ORDER BY department;

C. SELECT department, COUNT(*) FROM employees WHERE department;

D. SELECT department, COUNT(*) FROM employees HAVING department;

Solution

1. Step 1: Identify correct GROUP BY usage

   The GROUP BY clause must follow FROM and group by the column named.

2. Step 2: Check each option's clause

   SELECT department, COUNT(*) FROM employees GROUP BY department; uses GROUP BY correctly; others use ORDER BY, WHERE, HAVING incorrectly here.

3. Final Answer:

   SELECT department, COUNT(*) FROM employees GROUP BY department; -> Option A

4. Quick Check:

   GROUP BY syntax = SELECT ... GROUP BY column [OK]

Hint: GROUP BY follows FROM and groups by column [OK]

Common Mistakes:

• Using ORDER BY instead of GROUP BY
• Using WHERE to group data
• Using HAVING without aggregation

3. Given the table sales with columns region and amount, what is the output of this query?

SELECT region, SUM(amount) FROM sales GROUP BY region;

medium

A. A list of all sales amounts without grouping.

B. A list of regions with the total sales amount for each region.

C. An error because SUM() cannot be used with GROUP BY.

D. A list of regions sorted by amount.

Solution

1. Step 1: Understand GROUP BY with SUM()

   The query groups rows by region and sums the amount for each group.

2. Step 2: Predict output format

   The output shows each region once with the total amount of sales in that region.

3. Final Answer:

   A list of regions with the total sales amount for each region. -> Option B

4. Quick Check:

   GROUP BY + SUM() = grouped sums [OK]

Hint: GROUP BY + SUM() gives totals per group [OK]

Common Mistakes:

• Thinking SUM() can't be used with GROUP BY
• Expecting ungrouped list
• Confusing sorting with grouping

4. Identify the error in this SQL query:

SELECT department, COUNT(*) FROM employees;

medium

A. The query is correct and will run without errors.

B. COUNT(*) cannot be used without WHERE clause.

C. department cannot be selected without aggregation.

D. Missing GROUP BY clause for the department column.

Solution

1. Step 1: Check SELECT with aggregation

   COUNT(*) is an aggregate but department is not aggregated or grouped.

2. Step 2: Identify missing GROUP BY

   To select department with COUNT(*), GROUP BY department is required.

3. Final Answer:

   Missing GROUP BY clause for the department column. -> Option D

4. Quick Check:

   Non-aggregated columns need GROUP BY [OK]

Hint: Non-aggregated columns need GROUP BY [OK]

Common Mistakes:

• Ignoring missing GROUP BY
• Thinking COUNT(*) needs WHERE
• Assuming query runs without error

5. You have a table orders with columns customer_id, order_date, and total. You want to find the average order total per customer but only for customers who have placed more than 3 orders. Which query correctly achieves this?

hard

A. SELECT customer_id, AVG(total) FROM orders GROUP BY customer_id HAVING COUNT(*) > 3;

B. SELECT customer_id, AVG(total) FROM orders WHERE COUNT(*) > 3 GROUP BY customer_id;

C. SELECT customer_id, AVG(total) FROM orders GROUP BY customer_id WHERE COUNT(*) > 3;

D. SELECT customer_id, AVG(total) FROM orders HAVING COUNT(*) > 3 GROUP BY customer_id;

Solution

1. Step 1: Use GROUP BY to group orders by customer_id

   This groups all orders per customer to calculate aggregates.

2. Step 2: Use HAVING to filter groups with more than 3 orders

   HAVING filters groups after aggregation; WHERE cannot filter aggregates.

3. Final Answer:

   SELECT customer_id, AVG(total) FROM orders GROUP BY customer_id HAVING COUNT(*) > 3; -> Option A

4. Quick Check:

   HAVING filters groups, WHERE filters rows [OK]

Hint: Use HAVING to filter groups after GROUP BY [OK]

Common Mistakes:

• Using WHERE to filter aggregated counts
• Placing HAVING before GROUP BY
• Not filtering groups at all