Nhóm với GROUP BY

In Module 2 you learned that COUNT(*), AVG(...), SUM(...) reduce all rows to a single result row. But what if you want an average salary for each department, not for the whole company? You want one row per group. That's the job of GROUP BY.

The syntax

SELECT grouping_column, AGGREGATE_FUNCTION(...)
FROM   table
GROUP BY grouping_column;

Read in English: "split the rows into groups that share the same value of grouping_column, and for each group compute the aggregate".

-- Average salary per department:
SELECT department_id, AVG(salary) AS average
FROM   employees
GROUP BY department_id;

The result has one row for each distinct department_id found in employees. You'll get 5 rows: one for each of the 4 departments (1, 2, 3, 4) plus one for NULL (Andrea Costa).

With JOIN for readable names

Often you want to group by department_id but show the name of the department. You combine GROUP BY with a JOIN:

SELECT d.name AS department,
       COUNT(*)    AS count,
       AVG(e.salary) AS average_salary
FROM   employees   AS e
JOIN   departments AS d ON e.department_id = d.id
GROUP BY d.name
ORDER BY count DESC;

This is a full-fledged business intelligence query: in three lines it tells you how many employees each department has and how much they earn on average, sorted from largest to smallest.

The golden rule

When you use GROUP BY, every column in the SELECT must be:

1. one of the grouping columns (e.g. department_id, d.name), or
2. inside an aggregate function (e.g. AVG(salary), COUNT(*)).

-- ERROR: first_name is neither in GROUP BY nor inside an aggregate
SELECT department_id, first_name, AVG(salary)
FROM   employees
GROUP BY department_id;

PostgreSQL returns an explicit error ("column must appear in the GROUP BY clause or be used in an aggregate function"). It's a common early-weeks mistake: if you see it, stop and ask yourself "what is the value of first_name for a group of 3 people?" — there's no single answer, and that's why the engine refuses.

Try it

Review exercise