GROUP BY single column in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to group the sales by the product column.

SQL

SELECT product, SUM(amount) FROM sales [1] product;

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

AWHERE

BORDER BY

CGROUP BY

DHAVING

Attempts:

3 left

💡 Hint

Common Mistakes

Using ORDER BY instead of GROUP BY

Using WHERE to filter groups

The GROUP BY clause groups rows that have the same values in specified columns into summary rows.

2fill in blank

medium

Complete the code to find the total quantity sold for each category.

SQL

SELECT category, SUM(quantity) FROM products [1] category;

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

AHAVING

BGROUP BY

CORDER BY

DWHERE

Attempts:

3 left

💡 Hint

Common Mistakes

Using HAVING without GROUP BY

Using WHERE to filter groups

Use GROUP BY to group rows by category before summing quantities.

3fill in blank

hard

Fix the error in the query to correctly group by the customer_id column.

SQL

SELECT customer_id, COUNT(order_id) FROM orders [1] customer_id;

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

AGROUP BY

BWHERE

CJOIN

DORDER BY

Attempts:

3 left

💡 Hint

Common Mistakes

Using ORDER BY instead of GROUP BY

Missing GROUP BY clause

The query needs GROUP BY to group rows by customer_id before counting orders.

4fill in blank

hard

Fill both blanks to group sales by region and calculate the average sales amount.

SQL

SELECT [1], AVG(amount) FROM sales [2] region;

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

Aregion

BORDER BY

CGROUP BY

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Not including the grouping column in SELECT

Using ORDER BY instead of GROUP BY

The SELECT clause must include the grouping column region, and the query must use GROUP BY to group rows by region.

5fill in blank

hard

Fill all three blanks to select the department, count employees, and group by department.

SQL

SELECT [1], COUNT(employee_id) FROM employees [2] [3];

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

Adepartment

BGROUP BY

DORDER BY

Attempts:

3 left

💡 Hint

Common Mistakes

Using ORDER BY instead of GROUP BY

Not grouping by the selected column

The query selects department, counts employees, and groups by department using GROUP BY.

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