Multiple CTEs in one query in PostgreSQL

Introduction

Using multiple CTEs helps break down complex queries into smaller, easy-to-understand parts.

When you want to organize a big query into steps for clarity.

When you need to reuse intermediate results multiple times in one query.

When you want to improve readability by naming parts of your query.

When you want to separate different calculations before combining results.

Syntax

PostgreSQL

WITH cte1 AS (
  -- first query
),
cte2 AS (
  -- second query
)
SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id;

Each CTE is separated by a comma.

You can use the CTE names later in the main query.

Examples

This example uses two CTEs to get user names and their order totals, then joins them.

PostgreSQL

WITH first_cte AS (
  SELECT id, name FROM users
),
second_cte AS (
  SELECT user_id, order_total FROM orders
)
SELECT first_cte.name, second_cte.order_total
FROM first_cte
JOIN second_cte ON first_cte.id = second_cte.user_id;

This example creates a list of numbers and their squares using two CTEs.

PostgreSQL

WITH numbers AS (
  SELECT generate_series(1,5) AS num
),
squares AS (
  SELECT num, num * num AS square FROM numbers
)
SELECT * FROM squares;

Sample Program

This query uses two CTEs: one for employees and one for departments. Then it joins them to show each employee's department.

PostgreSQL

WITH employees AS (
  SELECT id, name, department_id FROM employee
),
departments AS (
  SELECT id, department_name FROM department
)
SELECT employees.name, departments.department_name
FROM employees
JOIN departments ON employees.department_id = departments.id
ORDER BY employees.name;

Important Notes

CTEs run before the main query and act like temporary tables.

Using multiple CTEs can make your SQL easier to read and maintain.

Summary

Multiple CTEs let you split complex queries into smaller parts.

Each CTE is named and separated by commas.

You can join or use CTEs in the final query to get combined results.

Practice

1. What is the main purpose of using multiple CTEs (Common Table Expressions) in a single PostgreSQL query?

easy

A. To speed up the database server hardware

B. To create multiple databases at once

C. To permanently store data in the database

D. To break a complex query into smaller, readable parts

Solution

1. Step 1: Understand what CTEs do

   CTEs let you write parts of a query separately with names, making complex queries easier to read.

2. Step 2: Identify the purpose of multiple CTEs

   Using multiple CTEs helps organize and simplify complex logic by splitting it into smaller named queries.

3. Final Answer:

   To break a complex query into smaller, readable parts -> Option D

4. Quick Check:

   Multiple CTEs = Query readability [OK]

Hint: Think of CTEs as named steps in a recipe [OK]

Common Mistakes:

• Confusing CTEs with permanent tables
• Thinking CTEs improve hardware speed
• Assuming CTEs create databases

2. Which of the following is the correct syntax to define two CTEs named cte1 and cte2 in one PostgreSQL query?

easy

A. WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id;

B. WITH cte1 (SELECT * FROM table1); WITH cte2 (SELECT * FROM table2); SELECT * FROM cte1, cte2;

C. WITH cte1 = SELECT * FROM table1, cte2 = SELECT * FROM table2 SELECT * FROM cte1 JOIN cte2;

D. WITH cte1 AS SELECT * FROM table1, cte2 AS SELECT * FROM table2 SELECT * FROM cte1 JOIN cte2;

Solution

1. Step 1: Recall correct CTE syntax

   CTEs start with WITH, each named, followed by AS and a query in parentheses, separated by commas.

2. Step 2: Check each option's syntax

   WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id; correctly uses commas, AS, and parentheses. Others miss parentheses or use wrong separators.

3. Final Answer:

   WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id; -> Option A

4. Quick Check:

   Correct CTE syntax = WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id; [OK]

Hint: Remember: WITH cte_name AS (query), separate multiple CTEs by commas [OK]

Common Mistakes:

• Omitting parentheses around CTE queries
• Using semicolons between CTEs
• Using '=' instead of AS

3. Given the following query, what will be the output?

WITH cte1 AS (SELECT 1 AS num), cte2 AS (SELECT num * 2 AS double_num FROM cte1) SELECT * FROM cte2;

medium

A. [{num: 2}]

B. [{num: 1, double_num: 2}]

C. [{double_num: 2}]

D. Syntax error

Solution

1. Step 1: Analyze cte1

   cte1 returns one row with column num = 1.

2. Step 2: Analyze cte2 using cte1

   cte2 selects num * 2 as double_num, so double_num = 1 * 2 = 2.

3. Step 3: Final SELECT from cte2

   Only double_num column is selected, so output is [{double_num: 2}].

4. Final Answer:

   [{double_num: 2}] -> Option C

5. Quick Check:

   cte2 output columns = double_num [OK]

Hint: Check which columns each CTE outputs carefully [OK]

Common Mistakes:

• Assuming cte2 includes all columns from cte1
• Expecting num column in final output
• Thinking query has syntax errors

4. Identify the error in this query using multiple CTEs:

WITH cte1 AS (SELECT id, name FROM users), cte2 AS (SELECT id, age FROM users) SELECT cte1.id, cte1.name, cte2.age FROM cte1 JOIN cte2 ON cte1.id = cte2.user_id;

medium

A. CTEs must be defined with semicolons

B. Column cte2.user_id does not exist

C. Missing comma between CTEs

D. Cannot join two CTEs in one query

Solution

1. Step 1: Check CTE definitions

   Both cte1 and cte2 select id and other columns from users table correctly.

2. Step 2: Check JOIN condition

   Query joins on cte1.id = cte2.user_id, but cte2 has no user_id column, only id.

3. Step 3: Identify error

   Using cte2.user_id causes a column not found error.

4. Final Answer:

   Column cte2.user_id does not exist -> Option B

5. Quick Check:

   Check column names in JOIN = Column cte2.user_id does not exist [OK]

Hint: Verify column names in JOIN conditions carefully [OK]

Common Mistakes:

• Assuming user_id exists in cte2
• Thinking semicolons separate CTEs
• Believing CTEs cannot be joined

5. You want to calculate the average order amount per customer and then find customers whose average order is above 100. Which query correctly uses multiple CTEs to achieve this?

hard

A. WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg;

B. WITH avg_orders AS (SELECT customer_id, SUM(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg;

C. WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg;

D. WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id) SELECT * FROM avg_orders WHERE avg_amount > 100;

Solution

1. Step 1: Calculate average order per customer

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; correctly uses AVG(amount) with GROUP BY customer_id in avg_orders CTE.

2. Step 2: Filter customers with avg_amount > 100

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; uses a second CTE high_avg to select customers meeting the condition.

3. Step 3: Final SELECT returns customers with high average orders

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; selects all from high_avg, giving correct result.

4. Final Answer:

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; -> Option A

5. Quick Check:

   Correct aggregation and filtering = WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; [OK]

Hint: Use one CTE for aggregation, another for filtering [OK]

Common Mistakes:

• Using SUM instead of AVG for average calculation
• Missing GROUP BY clause
• Not filtering in a separate CTE