GROUP BY multiple columns in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to group the sales by both city and product.

SQL

SELECT city, product, SUM(sales) FROM sales_data GROUP BY [1];

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

Acity, product

Bcity

Cproduct

Dsales

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by only one column when multiple are needed.

Including aggregated columns in GROUP BY.

The GROUP BY clause must include both city and product to group by multiple columns.

2fill in blank

medium

Complete the code to count the number of orders grouped by customer and order_date.

SQL

SELECT customer_id, order_date, COUNT(*) FROM orders GROUP BY [1];

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

Acustomer_id

Bcustomer_id, order_date

Corder_date

Dorder_id

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by only one column when two are needed.

Grouping by a column not selected.

To group by both customer_id and order_date, list both columns separated by a comma in the GROUP BY clause.

3fill in blank

hard

Fix the error in the GROUP BY clause to correctly group by both department and role.

SQL

SELECT department, role, AVG(salary) FROM employees GROUP BY [1];

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

Adepartment role

Bsalary

Crole, department,

Ddepartment, role

Attempts:

3 left

💡 Hint

Common Mistakes

Missing commas between columns.

Including aggregated columns in GROUP BY.

The columns in GROUP BY must be separated by commas without extra trailing commas.

4fill in blank

hard

Fill both blanks to group sales by region and product category.

SQL

SELECT [1], [2], SUM(amount) FROM sales GROUP BY [1], [2];

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

Aregion

Bamount

Cproduct_category

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by columns not selected.

Using aggregated columns in GROUP BY.

You must select and group by the same columns: region and product_category.

5fill in blank

hard

Fill all three blanks to select city, category, and total sales grouped by city and category.

SQL

SELECT [1], [2], SUM([3]) FROM sales_data GROUP BY [1], [2];

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

Acity

Bcategory

Csales

Ddate

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by columns not in SELECT.

Using wrong column inside SUM().

The query selects city and category and sums the sales column, grouping by city and category.

Practice

1. What does the SQL clause GROUP BY column1, column2 do?

easy

A. Sorts the table by column1 and then column2

B. Groups rows by unique combinations of values in column1 and column2

C. Filters rows where column1 equals column2

D. Joins two tables on column1 and column2

Solution

1. Step 1: Understand GROUP BY purpose

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

2. Step 2: Apply to multiple columns

   When multiple columns are listed, grouping happens on unique combinations of those columns' values.

3. Final Answer:

   Groups rows by unique combinations of values in column1 and column2 -> Option B

4. Quick Check:

   GROUP BY multiple columns = group by combinations [OK]

Hint: GROUP BY multiple columns groups by combined unique values [OK]

Common Mistakes:

• Thinking GROUP BY sorts data
• Confusing GROUP BY with WHERE filtering
• Assuming GROUP BY joins tables

2. Which of the following is the correct syntax to group data by two columns named city and year?

easy

A. SELECT city, year FROM table GROUP city, year;

B. SELECT city, year FROM table ORDER BY city, year;

C. SELECT city, year FROM table GROUP BY city, year;

D. SELECT city, year FROM table GROUP BY city year;

Solution

1. Step 1: Recall GROUP BY syntax

   The correct syntax is GROUP BY followed by column names separated by commas.

2. Step 2: Check each option

   SELECT city, year FROM table GROUP BY city, year; uses correct syntax with commas. SELECT city, year FROM table ORDER BY city, year; uses ORDER BY, which is for sorting. SELECT city, year FROM table GROUP city, year; misses BY keyword. SELECT city, year FROM table GROUP BY city year; misses comma between columns.

3. Final Answer:

   SELECT city, year FROM table GROUP BY city, year; -> Option C

4. Quick Check:

   GROUP BY columns separated by commas [OK]

Hint: Use GROUP BY with commas between columns [OK]

Common Mistakes:

• Using ORDER BY instead of GROUP BY
• Omitting BY keyword after GROUP
• Missing commas between column names

3. Given the table sales with columns region, product, and amount, what will this query return?

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

medium

A. Total sales amount for each region only

B. Syntax error due to missing GROUP BY columns

C. Total sales amount for each product only

D. Total sales amount for each region and product combination

Solution

1. Step 1: Analyze SELECT and GROUP BY columns

   The query groups rows by both region and product, so each group is a unique pair of region and product.

2. Step 2: Understand aggregation function SUM(amount)

   SUM(amount) calculates total sales amount for each group of region and product.

3. Final Answer:

   Total sales amount for each region and product combination -> Option D

4. Quick Check:

   GROUP BY region, product + SUM = totals per pair [OK]

Hint: GROUP BY columns + SUM aggregates per group [OK]

Common Mistakes:

• Thinking it sums only by one column
• Assuming syntax error without reason
• Ignoring that all selected non-aggregated columns must be grouped

4. Identify the error in this query:

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

medium

A. Missing role column in GROUP BY clause

B. COUNT(*) cannot be used with GROUP BY

C. department should not be in GROUP BY

D. SELECT must include only aggregated columns

Solution

1. Step 1: Check SELECT columns vs GROUP BY columns

   Columns in SELECT that are not aggregated must appear in GROUP BY. Here, role is in SELECT but missing in GROUP BY.

2. Step 2: Understand aggregation rules

   COUNT(*) is valid, but all non-aggregated columns must be grouped to avoid errors.

3. Final Answer:

   Missing role column in GROUP BY clause -> Option A

4. Quick Check:

   All non-aggregated SELECT columns must be in GROUP BY [OK]

Hint: Include all non-aggregated SELECT columns in GROUP BY [OK]

Common Mistakes:

• Ignoring missing columns in GROUP BY
• Thinking COUNT(*) is invalid with GROUP BY
• Assuming GROUP BY only needs one column

5. You have a transactions table with columns customer_id, month, and amount. You want to find the average transaction amount per customer per month, but only for months where the customer made more than 3 transactions. Which query correctly achieves this?

hard

A. SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY customer_id, month HAVING COUNT(*) > 3;

B. SELECT customer_id, month, AVG(amount) FROM transactions WHERE COUNT(*) > 3 GROUP BY customer_id, month;

C. SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY customer_id HAVING COUNT(*) > 3;

D. SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY month HAVING COUNT(*) > 3;

Solution

1. Step 1: Understand filtering groups with HAVING

   HAVING filters groups after grouping. To filter groups with more than 3 transactions, use HAVING COUNT(*) > 3.

2. Step 2: Group by both customer_id and month

   To get average per customer per month, group by both columns.

3. Step 3: Check each option

   SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY customer_id, month HAVING COUNT(*) > 3; correctly uses GROUP BY customer_id, month and HAVING COUNT(*) > 3. SELECT customer_id, month, AVG(amount) FROM transactions WHERE COUNT(*) > 3 GROUP BY customer_id, month; misuses WHERE with COUNT(). SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY customer_id HAVING COUNT(*) > 3; groups only by customer_id, missing month. SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY month HAVING COUNT(*) > 3; groups only by month, missing customer_id.

4. Final Answer:

   SELECT customer_id, month, AVG(amount) FROM transactions GROUP BY customer_id, month HAVING COUNT(*) > 3; -> Option A

5. Quick Check:

   Use HAVING to filter grouped counts [OK]

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

Common Mistakes:

• Using WHERE with aggregate functions
• Grouping by only one column when two needed
• Filtering before grouping instead of after