1. JOIN or INNER JOIN
2. OUTER JOIN
2.1 LEFT OUTER JOIN or LEFT JOIN
2.2 RIGHT OUTER JOIN or RIGHT JOIN
2.3 FULL OUTER JOIN or FULL JOIN
3. NATURAL JOIN
4. CROSS JOIN
5. SELF JOIN
1. 连接或内连接:
在这种JOIN类型中,我们获得两个表中匹配条件的所有记录,而两个表中不匹配的记录将不被报告。
换句话说,INNER JOIN基于这样一个事实:只有两个表中匹配的条目才应该被列出。
注意,没有任何其他JOIN关键字(如INNER, OUTER, LEFT等)的JOIN是INNER JOIN。换句话说,JOIN是
a INNER JOIN的语法糖(参见:连接和内部连接的区别)
SELECT *
-- This just generates all the days in January 2017
FROM generate_series(
'2017-01-01'::TIMESTAMP,
'2017-01-01'::TIMESTAMP + INTERVAL '1 month -1 day',
INTERVAL '1 day'
) AS days(day)
-- Here, we're combining all days with all departments
CROSS JOIN departments
它将一个表中的所有行与另一个表中的所有行组合在一起:
来源:
+--------+ +------------+
| day | | department |
+--------+ +------------+
| Jan 01 | | Dept 1 |
| Jan 02 | | Dept 2 |
| ... | | Dept 3 |
| Jan 30 | +------------+
| Jan 31 |
+--------+
结果:
+--------+------------+
| day | department |
+--------+------------+
| Jan 01 | Dept 1 |
| Jan 01 | Dept 2 |
| Jan 01 | Dept 3 |
| Jan 02 | Dept 1 |
| Jan 02 | Dept 2 |
| Jan 02 | Dept 3 |
| ... | ... |
| Jan 31 | Dept 1 |
| Jan 31 | Dept 2 |
| Jan 31 | Dept 3 |
+--------+------------+
如果我们只是写一个逗号分隔的表列表,我们会得到相同的结果:
-- CROSS JOINing two tables:
SELECT * FROM table1, table2
SELECT *
-- Same as before
FROM generate_series(
'2017-01-01'::TIMESTAMP,
'2017-01-01'::TIMESTAMP + INTERVAL '1 month -1 day',
INTERVAL '1 day'
) AS days(day)
-- Now, exclude all days/departments combinations for
-- days before the department was created
JOIN departments AS d ON day >= d.created_at
-- Convenient syntax:
SELECT *
FROM a LEFT JOIN b ON <predicate>
-- Cumbersome, equivalent syntax:
SELECT a.*, b.*
FROM a JOIN b ON <predicate>
UNION ALL
SELECT a.*, NULL, NULL, ..., NULL
FROM a
WHERE NOT EXISTS (
SELECT * FROM b WHERE <predicate>
)
-- Oracle
SELECT *
FROM actor a, film_actor fa, film f
WHERE a.actor_id = fa.actor_id(+)
AND fa.film_id = f.film_id(+)
-- SQL Server
SELECT *
FROM actor a, film_actor fa, film f
WHERE a.actor_id *= fa.actor_id
AND fa.film_id *= f.film_id
WITH
-- Using CONNECT BY to generate all dates in January
days(day) AS (
SELECT DATE '2017-01-01' + LEVEL - 1
FROM dual
CONNECT BY LEVEL <= 31
),
-- Our departments
departments(department, created_at) AS (
SELECT 'Dept 1', DATE '2017-01-10' FROM dual UNION ALL
SELECT 'Dept 2', DATE '2017-01-11' FROM dual UNION ALL
SELECT 'Dept 3', DATE '2017-01-12' FROM dual UNION ALL
SELECT 'Dept 4', DATE '2017-04-01' FROM dual UNION ALL
SELECT 'Dept 5', DATE '2017-04-02' FROM dual
)
SELECT *
FROM days
LEFT JOIN departments
PARTITION BY (department) -- This is where the magic happens
ON day >= created_at
结果如下:
+--------+------------+------------+
| day | department | created_at |
+--------+------------+------------+
| Jan 01 | Dept 1 | | -- Didn't match, but still get row
| Jan 02 | Dept 1 | | -- Didn't match, but still get row
| ... | Dept 1 | | -- Didn't match, but still get row
| Jan 09 | Dept 1 | | -- Didn't match, but still get row
| Jan 10 | Dept 1 | Jan 10 | -- Matches, so get join result
| Jan 11 | Dept 1 | Jan 10 | -- Matches, so get join result
| Jan 12 | Dept 1 | Jan 10 | -- Matches, so get join result
| ... | Dept 1 | Jan 10 | -- Matches, so get join result
| Jan 31 | Dept 1 | Jan 10 | -- Matches, so get join result
SELECT *
FROM actor a
WHERE NOT EXISTS (
SELECT * FROM film_actor fa
WHERE a.actor_id = fa.actor_id
)
有些人(尤其是MySQL的人)也会这样写ANTI - JOIN:
SELECT *
FROM actor a
LEFT JOIN film_actor fa
USING (actor_id)
WHERE film_id IS NULL
我认为历史原因是表现。
横向连接
天哪,这个太酷了。只有我一个人提起这件事?这是一个很酷的问题:
SELECT a.first_name, a.last_name, f.*
FROM actor AS a
LEFT OUTER JOIN LATERAL (
SELECT f.title, SUM(amount) AS revenue
FROM film AS f
JOIN film_actor AS fa USING (film_id)
JOIN inventory AS i USING (film_id)
JOIN rental AS r USING (inventory_id)
JOIN payment AS p USING (rental_id)
WHERE fa.actor_id = a.actor_id -- JOIN predicate with the outer query!
GROUP BY f.film_id
ORDER BY revenue DESC
LIMIT 5
) AS f
ON true
SELECT a.first_name, a.last_name, f.*
FROM actor AS a
OUTER APPLY (
SELECT f.title, SUM(amount) AS revenue
FROM film AS f
JOIN film_actor AS fa ON f.film_id = fa.film_id
JOIN inventory AS i ON f.film_id = i.film_id
JOIN rental AS r ON i.inventory_id = r.inventory_id
JOIN payment AS p ON r.rental_id = p.rental_id
WHERE fa.actor_id = a.actor_id -- JOIN predicate with the outer query!
GROUP BY f.film_id
ORDER BY revenue DESC
LIMIT 5
) AS f