Why aggregation matters in PostgreSQL - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to count the number of orders in the orders table.

PostgreSQL

SELECT [1](order_id) FROM orders;

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

AMIN

BSUM

CMAX

DCOUNT

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM instead of COUNT will try to add order IDs, which may not make sense.

Using MAX or MIN returns the highest or lowest order ID, not the count.

The COUNT function counts the number of rows or non-null values in a column. Here, it counts all order IDs.

2fill in blank

medium

Complete the code to find the average price from the products table.

PostgreSQL

SELECT [1](price) FROM products;

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

AAVG

BMIN

CSUM

DCOUNT

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM returns the total price, not the average.

Using COUNT returns the number of prices, not their average.

The AVG function calculates the average (mean) of numeric values in a column.

3fill in blank

hard

Fix the error in the query to get the total quantity sold from sales.

PostgreSQL

SELECT [1](quantity) FROM sales;

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

ACOUNT

BSUM

CAVG

DMAX

Attempts:

3 left

💡 Hint

Common Mistakes

Using COUNT counts rows, not the total quantity.

Using AVG returns the average quantity, not the total.

The SUM function adds all values in the quantity column to get the total sold.

4fill in blank

hard

Fill both blanks to get the number of orders per customer.

PostgreSQL

SELECT customer_id, [1](order_id) FROM orders GROUP BY [2];

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

ACOUNT

BSUM

Ccustomer_id

Dorder_id

Attempts:

3 left

💡 Hint

Common Mistakes

Grouping by order_id would group each order separately.

Using SUM instead of COUNT counts values, not rows.

We count orders per customer, so use COUNT(order_id) and group by customer_id.

5fill in blank

hard

Fill all three blanks to find the average price per category for products priced above 100.

PostgreSQL

SELECT [1], [2](price) FROM products WHERE price [3] 100 GROUP BY category;

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

Acategory

BAVG

C>

DSUM

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM instead of AVG calculates total price, not average.

Using < instead of > filters prices below 100.

We select the category, calculate the average price with AVG(price), filter prices greater than 100, and group by category.

Practice

1. What is the main purpose of aggregation functions like COUNT or SUM in SQL?

easy

A. To summarize multiple rows into a single value

B. To delete duplicate rows from a table

C. To join two tables together

D. To create new tables from existing ones

Solution

1. Step 1: Understand aggregation functions

   Aggregation functions like COUNT and SUM take many rows and produce one summary value.

2. Step 2: Identify their purpose

   They help to reduce data by summarizing, not by deleting or joining tables.

3. Final Answer:

   To summarize multiple rows into a single value -> Option A

4. Quick Check:

   Aggregation = Summary [OK]

Hint: Aggregation means summarizing many rows into one value [OK]

Common Mistakes:

• Thinking aggregation deletes rows
• Confusing aggregation with JOIN
• Believing aggregation creates new tables

2. Which of the following is the correct syntax to count rows in a table named sales?

easy

A. SELECT COUNT ALL sales;

B. SELECT COUNT(*) FROM sales;

C. SELECT COUNT(sales) FROM *;

D. SELECT COUNT FROM sales();

Solution

1. Step 1: Recall correct COUNT syntax

   The correct syntax to count all rows is SELECT COUNT(*) FROM table_name;.

2. Step 2: Check each option

   SELECT COUNT(*) FROM sales; matches the correct syntax. Others have syntax errors or wrong order.

3. Final Answer:

   SELECT COUNT(*) FROM sales; -> Option B

4. Quick Check:

   COUNT(*) syntax = SELECT COUNT(*) FROM sales; [OK]

Hint: COUNT(*) counts all rows, use parentheses and star [OK]

Common Mistakes:

• Omitting parentheses after COUNT
• Using COUNT without * or column
• Wrong order of keywords

3. Given the table orders with columns customer_id and amount, what does this query return?

SELECT customer_id, SUM(amount) FROM orders GROUP BY customer_id;

medium

A. Total amount spent by each customer

B. List of all orders with amounts

C. Total amount spent by all customers combined

D. Number of orders per customer

Solution

1. Step 1: Understand GROUP BY usage

   GROUP BY customer_id groups rows by each customer.

2. Step 2: Understand SUM(amount)

   SUM(amount) adds all amounts per customer group.

3. Final Answer:

   Total amount spent by each customer -> Option A

4. Quick Check:

   GROUP BY + SUM = total per group [OK]

Hint: GROUP BY groups rows; SUM adds values per group [OK]

Common Mistakes:

• Thinking SUM returns total for all rows without grouping
• Confusing SUM with COUNT
• Ignoring GROUP BY effect

4. What is wrong with this query?

SELECT customer_id, COUNT(*) FROM orders;

medium

A. customer_id must be inside COUNT()

B. COUNT(*) cannot be used without WHERE

C. Missing GROUP BY clause for customer_id

D. SELECT must have ORDER BY with COUNT

Solution

1. Step 1: Check aggregation with non-aggregated column

   customer_id is selected but not aggregated or grouped.

2. Step 2: Identify missing GROUP BY

   To use COUNT(*) with customer_id, GROUP BY customer_id is required.

3. Final Answer:

   Missing GROUP BY clause for customer_id -> Option C

4. Quick Check:

   Non-aggregated columns need GROUP BY [OK]

Hint: Use GROUP BY when mixing columns with aggregation [OK]

Common Mistakes:

• Forgetting GROUP BY when selecting non-aggregated columns
• Thinking COUNT needs WHERE clause
• Assuming ORDER BY is mandatory with aggregation

5. You want to find the average order amount per customer but only for customers who placed more than 3 orders. Which query correctly achieves this?

hard

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

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

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

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

Solution

1. Step 1: Understand filtering groups with HAVING

   HAVING filters groups after aggregation; WHERE filters rows before aggregation.

2. Step 2: Check query order and clauses

   SELECT customer_id, AVG(amount) FROM orders GROUP BY customer_id HAVING COUNT(*) > 3; correctly uses GROUP BY then HAVING COUNT(*) > 3 to filter customers with more than 3 orders.

3. Final Answer:

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

4. Quick Check:

   Use HAVING to filter groups after GROUP BY [OK]

Hint: Use HAVING to filter groups, not WHERE [OK]

Common Mistakes:

• Using WHERE to filter aggregated counts
• Placing HAVING before GROUP BY
• Syntax errors mixing WHERE and HAVING