LeetCode（数据库）- 周内每天的销售情况

Posted 2021-12-19 程序员牧码

题目链接：点击打开链接

题目大意：略。

解题思路：注意 DAYOFWEEK 函数。

AC 代码

-- 解决方案(1)
select distinct b.item_category as Category,
ifnull(sum(case when dayofweek(a.order_date) = 2 then a.quantity end),0) Monday,
ifnull(sum(case when dayofweek(a.order_date) = 3 then a.quantity end),0) Tuesday,
ifnull(sum(case when dayofweek(a.order_date) = 4 then a.quantity end),0) Wednesday,
ifnull(sum(case when dayofweek(a.order_date) = 5 then a.quantity end),0) Thursday,
ifnull(sum(case when dayofweek(a.order_date) = 6 then a.quantity end),0) Friday,
ifnull(sum(case when dayofweek(a.order_date) = 7 then a.quantity end),0) Saturday,
ifnull(sum(case when dayofweek(a.order_date) = 1 then a.quantity end),0) Sunday
from Orders a right join Items b
on a.item_id = b.item_id
group by Category
order by Category

-- 解决方案(2)
WITH t AS(SELECT item_category, order_date, quantity, DAYOFWEEK(order_date) - 1 week FROM Orders JOIN Items USING(item_id)),

tt AS(SELECT *, SUM(quantity) total FROM t GROUP BY item_category, week),

ttt AS(SELECT item_category, 
IF(week = 1, total, 0) Monday,
IF(week = 2, total, 0) Tuesday,
IF(week = 3, total, 0) Wednesday,
IF(week = 4, total, 0) Thursday, 
IF(week = 5, total, 0) Friday,
IF(week = 6, total, 0) Saturday, 
IF(week = 0, total, 0) Sunday
FROM tt)

SELECT DISTINCT item_category Category, IFNULL(SUM(Monday), 0) Monday, IFNULL(SUM(Tuesday), 0) Tuesday, IFNULL(SUM(Wednesday), 0) Wednesday, IFNULL(SUM(Thursday), 0) Thursday, IFNULL(SUM(Friday), 0) Friday, IFNULL(SUM(Saturday), 0) Saturday, IFNULL(SUM(Sunday), 0) Sunday
FROM ttt RIGHT JOIN (SELECT DISTINCT item_category FROM Items) tttt USING(item_category)
GROUP BY item_category
ORDER BY item_category