[SQL]1045(JOIN)+603(abs, JOIN)

1045. 买下全部产品的客户

思路

1. 在Customer表中计算每一个客户买Product的种类，为防重复购买，COUNT()要配合DISTINCT食用。建立临时表a。

FROM (SELECT customer_id, COUNT(DISTINCT product_key) AS num FROM Customer GROUP BY customer_id) a

2. 计算Product表中的类型数。建立临时表b。

(SELECT COUNT(product_key) AS num FROM Product) b

3.用自联结(JOIN取交集)

SELECT customer_id
FROM (SELECT customer_id, COUNT(DISTINCT product_key) AS num FROM Customer GROUP BY customer_id) a
JOIN
(SELECT COUNT(product_key) AS num FROM Product) b 
ON a.num = b.num;

603. 连续空余座位

自链接

select a.seat_id, a.free, b.seat_id, b.free
from cinema a join cinema b;

找到两个连续的空的座位

select a.seat_id, a.free, b.seat_id, b.free
from cinema a join cinema b
  on abs(a.seat_id - b.seat_id) = 1
  and a.free = true and b.free = true;

最终代码

SELECT DISTINCT a.seat_id
FROM cinema a JOIN cinema b 
ON abs(a.seat_id - b.seat_id) = 1
AND a.free = TRUE AND b.free = TRUE
ORDER BY a.seat_id;