PostgreSQL DISTINCT ON

2017-05-02 ·

DISTINCT clause eliminates duplicate rows from the results retrieved by SELECT statement. It keeps one row for each group of duplicates. This one row is unpredictable unless ORDER BY is used to ensure that the desired row appears first

SELECT DISTINCT department FROM employees;

DISTINCT can be also used on multiple columns at once; in that case it will evaluate the duplicates based on the combination of values of those columns.

DISTINCT behavior can be simulated by GROUP BY clause.

SELECT department FROM employees GROUP BY department;

Let's imagine we have a two separate tables: users and logins. We will use DISTINCT ON to find the information about the most recent login by each user.

CREATE TABLE users (
	username text not null,
	email text not null,
  CONSTRAINT users_pk PRIMARY KEY (username)
);
INSERT INTO users (username, email) VALUES
 ('Zosia', 'zosia@example.com'),
 ('Jasiu', 'jasiu@example.com'),
 ('Krysia', 'krysia@example.com');
CREATE TABLE logins (
  username text not null,
	browser text not null,
	logged_at timestamp not null,
  CONSTRAINT logins_pk PRIMARY KEY (username, logged_at),
  CONSTRAINT logins_fk_username FOREIGN KEY (username) REFERENCES users(username)
);
INSERT INTO logins (username, browser, logged_at) VALUES
  ('Zosia', 'Safari', '2000-01-01 11:10:01'),
  ('Zosia', 'Chrome', '2000-01-02 11:10:01'),
  ('Zosia', 'Safari', '2000-01-03 11:10:01'),
  ('Jasiu', 'Firefox', '2000-02-01 11:10:01'),
  ('Jasiu', 'Chrome', '2000-02-02 11:10:01'),
  ('Jasiu', 'Chrome', '2000-02-03 11:10:01'),
  ('Krysia', 'Opera', '2000-03-01 11:10:01'),
  ('Krysia', 'Safari', '2000-03-02 11:10:01'),
  ('Krysia', 'Opera', '2000-03-03 11:10:01');

Here's our query which joins those two tables and prints the most recent login. DISTINCT ON requires that its expression (i.e. u.username in our example) must match the first expression used in ORDER BY clause.

SELECT DISTINCT ON (u.username) u.username, u.email, l.browser, l.logged_at
FROM users u
JOIN logins l ON l.username = u.username
ORDER BY u.username, logged_at DESC

Here's the result:

 username │       email        │ browser │      logged_at
──────────┼────────────────────┼─────────┼─────────────────────
 Jasiu    │ jasiu@example.com  │ Chrome  │ 2000-02-03 11:10:01
 Krysia   │ krysia@example.com │ Opera   │ 2000-03-03 11:10:01
 Zosia    │ zosia@example.com  │ Safari  │ 2000-01-03 11:10:01

Let's consider another example: /find the employee with the highest salary in each department/. The naive solution could look as the following:

SELECT  * FROM employees WHERE
  (department, salary) IN (
    SELECT department, MAX(salary) FROM employees
    GROUP BY department
  )
ORDER BY department;

We could also use PARTITION to achieve the same effect.

WITH ranked_employees AS (
  SELECT ROW_NUMBER() OVER (
    PARTITION BY department ORDER BY salary DESC
  ) AS rn, * FROM employees
)
SELECT * FROM ranked_employees
WHERE rn = 1
ORDER BY department;

PostgreSQL's DISTINCT ON clause, however, greatly simplifies that query.

SELECT DISTINCT ON (department) * FROM employees
ORDER BY department, salary DESC;
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