SQLで重複を見つける方法ガイド

SELECT columns, COUNT(*)
FROM table_name
GROUP BY columns
HAVING COUNT(*) > 1;

-- Input: customers
-- id | name           | email
-- ---+----------------+----------------------
-- 1  | Maria Lopez    | maria.lopez@gmail.com
-- 2  | Sam Patel      | sam@patel.dev
-- 3  | M. Lopez       | maria.lopez@gmail.com
-- 4  | Kenji Watanabe | kenji@watanabe.jp

SELECT email, COUNT(*) AS copies
FROM customers
GROUP BY email
HAVING COUNT(*) > 1;

-- Output:
-- email                 | copies
-- ----------------------+-------
-- maria.lopez@gmail.com | 2

-- Input: the customers table above

SELECT name, email, COUNT(*) AS copies
FROM customers
GROUP BY email
HAVING COUNT(*) > 1;

-- ERROR: column "customers.name" must appear in the GROUP BY clause or be used in an aggregate function

-- Input: users
-- id | email
-- ---+--------------
-- 1  | ana@silva.dev
-- 2  | NULL
-- 3  | Ana@Silva.dev
-- 4  | NULL

SELECT email, COUNT(*) AS copies
FROM users
GROUP BY email
HAVING COUNT(*) > 1;

-- Output:
-- email | copies
-- ------+-------
-- NULL  | 2

-- MySQL and SQL Server ignore case by default, so they also return ana@silva.dev | 2

-- Skip the NULLs and ignore case:
SELECT LOWER(email) AS email, COUNT(*) AS copies
FROM users
WHERE email IS NOT NULL
GROUP BY LOWER(email)
HAVING COUNT(*) > 1;

-- Output:
-- email         | copies
-- --------------+-------
-- ana@silva.dev | 2

-- Input: the customers table above

SELECT id, name, email, copies
FROM (
  SELECT id, name, email, COUNT(*) OVER (PARTITION BY email) AS copies
  FROM customers
) AS c
WHERE copies > 1
ORDER BY id;

-- Output:
-- id | name        | email                 | copies
-- ---+-------------+-----------------------+-------
-- 1  | Maria Lopez | maria.lopez@gmail.com | 2
-- 3  | M. Lopez    | maria.lopez@gmail.com | 2

-- Input: the customers table above

SELECT id, name, email
FROM customers c
WHERE EXISTS (
  SELECT 1 FROM customers other
  WHERE other.email = c.email AND other.id <> c.id
);

-- Output:
-- id | name        | email
-- ---+-------------+----------------------
-- 1  | Maria Lopez | maria.lopez@gmail.com
-- 3  | M. Lopez    | maria.lopez@gmail.com

-- Input: bookings
-- booking_id | room | night      | guest
-- -----------+------+------------+----------
-- 1          | 204  | 2026-11-06 | R. Okafor
-- 2          | 204  | 2026-11-07 | R. Okafor
-- 3          | 310  | 2026-11-06 | L. Chen
-- 4          | 204  | 2026-11-06 | P. Novak

SELECT room, night, COUNT(*) AS bookings
FROM bookings
GROUP BY room, night
HAVING COUNT(*) > 1;

-- Output:
-- room | night      | bookings
-- -----+------------+---------
-- 204  | 2026-11-06 | 2

-- Input: accounts
-- id | username | email
-- ---+----------+----------------
-- 1  | jkim     | jin@kim.kr
-- 2  | olivia_r | olivia@reyes.mx
-- 3  | jin.kim  | Jin@Kim.kr
-- 4  | noah_b   | noah@baker.uk

SELECT id, username, email, accounts_with_email
FROM (
  SELECT id, username, email,
         COUNT(*) OVER (PARTITION BY LOWER(email)) AS accounts_with_email
  FROM accounts
) AS a
WHERE accounts_with_email > 1
ORDER BY id;

-- Output:
-- id | username | email      | accounts_with_email
-- ---+----------+------------+--------------------
-- 1  | jkim     | jin@kim.kr | 2
-- 3  | jin.kim  | Jin@Kim.kr | 2

-- Input: users
-- id | first_name | last_name | email
-- ---+------------+-----------+----------------
-- 1  | Priya      | Shah      | priya@shah.net
-- 2  | Tom        | Becker    | tom@becker.de
-- 3  | Priya      | Shah      | pshah@work.com
-- 4  | Lena       | Kowalski  | lena@kowal.pl
-- 5  | Tom        | Becker    | tbecker@mail.de

WITH numbered AS (
  SELECT id, first_name, last_name, email,
         ROW_NUMBER() OVER (PARTITION BY first_name, last_name ORDER BY id) AS copy_number
  FROM users
)
SELECT id, first_name, last_name, email,
       CASE WHEN copy_number > 1 THEN 'Duplicate' ELSE 'Original' END AS status
FROM numbered
ORDER BY id;

-- Output:
-- id | first_name | last_name | email           | status
-- ---+------------+-----------+-----------------+----------
-- 1  | Priya      | Shah      | priya@shah.net  | Original
-- 2  | Tom        | Becker    | tom@becker.de   | Original
-- 3  | Priya      | Shah      | pshah@work.com  | Duplicate
-- 4  | Lena       | Kowalski  | lena@kowal.pl   | Original
-- 5  | Tom        | Becker    | tbecker@mail.de | Duplicate

-- Input: orders
-- order_id | customer_id | product   | order_date
-- ---------+-------------+-----------+-----------
-- 1001     | 42          | Desk Lamp | 2026-03-02
-- 1002     | 42          | Desk Lamp | 2026-03-02
-- 1003     | 42          | Desk Lamp | 2026-03-09
-- 1004     | 57          | Desk Lamp | 2026-03-02

SELECT o1.order_id AS first_order, o2.order_id AS repeat_order,
       o1.customer_id, o1.product
FROM orders o1
JOIN orders o2
  ON o1.customer_id = o2.customer_id
 AND o1.product = o2.product
 AND o1.order_date = o2.order_date
 AND o1.order_id < o2.order_id;

-- Output:
-- first_order | repeat_order | customer_id | product
-- ------------+--------------+-------------+----------
-- 1001        | 1002         | 42          | Desk Lamp