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Intermediate SQL

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Using Inner Subqueries in the FROM Section

An inner subquery creates a table with which we then perform operations in the outer query.

From this, we can conclude that we can also use an inner query within the FROM section to select something from this table.

Let's consider an example where we need to work with employees whose salary is above a certain value. But instead of a WHERE clause, we will use an inner query:

SELECT *
FROM (
SELECT *
From employees
WHERE salary >= 70000
) AS high_salary_employees

              
123456

            
SELECT *
FROM (
  SELECT * 
  From employees
  WHERE salary >= 70000
) AS high_salary_employees
copy

We've created a new table using an inner subquery with which we can work and gather information.

Note

Notice that for such a table, it's essential to provide an alias.

When might this be useful?

Such a query is highly readable; sometimes, it can replace a WHERE clause.

Additionally, if you need to work not with the entire table but only with a filtered part, an inner subquery within the FROM section can be extremely handy.

Task

Swipe to start coding

Your task is to retrieve the average budget (budget column) for all managerial departments from the department table. Use an inner subquery in the FROM section to filter only the managerial type of department. Also, use the alias manager_departments for this table so the task is checked correctly.

The response should have only one column, average_budget. This is an alias for the aggregate function AVG().

Note

This task could also be done using a WHERE clause, but we use a nested query here to practice this syntax, which will be useful for solving more complex problems in the future.

Brief Instructions

  • Retrieve the average value of the budget column.
  • Assign the alias average_budget to that column.
  • In the FROM clause, you need to write a nested query.
  • In the nested query, select all columns from the department table.
  • In the nested query, use the WHERE clause with the condition type = 'manager'.
  • Assign the alias manager_departments to the nested query.

Solution

SELECT AVG(budget) as average_budget
FROM (
SELECT *
FROM department
WHERE type = 'manager'
) AS manager_departments

Everything was clear?

How can we improve it?

Thanks for your feedback!

Section 2. Chapter 3

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