# Write your MySQL query statement below
1479. 周内每天的销售情况
表:Orders
+---------------+---------+ | Column Name | Type | +---------------+---------+ | order_id | int | | customer_id | int | | order_date | date | | item_id | varchar | | quantity | int | +---------------+---------+ (order_id, item_id) 是该表主键(具有唯一值的列的组合) 该表包含了订单信息 order_date 是id为 item_id 的商品被id为 customer_id 的消费者订购的日期.
表:Items
+---------------------+---------+ | Column Name | Type | +---------------------+---------+ | item_id | varchar | | item_name | varchar | | item_category | varchar | +---------------------+---------+ item_id 是该表主键(具有唯一值的列) item_name 是商品的名字 item_category 是商品的类别
你是企业主,想要获得分类商品和周内每天的销售报告。
编写解决方案,报告 周内每天 每个商品类别下订购了多少单位。
返回结果表单 按商品类别排序 。
结果格式如下例所示。
示例 1:
输入: Orders
表: +------------+--------------+-------------+--------------+-------------+ | order_id | customer_id | order_date | item_id | quantity | +------------+--------------+-------------+--------------+-------------+ | 1 | 1 | 2020-06-01 | 1 | 10 | | 2 | 1 | 2020-06-08 | 2 | 10 | | 3 | 2 | 2020-06-02 | 1 | 5 | | 4 | 3 | 2020-06-03 | 3 | 5 | | 5 | 4 | 2020-06-04 | 4 | 1 | | 6 | 4 | 2020-06-05 | 5 | 5 | | 7 | 5 | 2020-06-05 | 1 | 10 | | 8 | 5 | 2020-06-14 | 4 | 5 | | 9 | 5 | 2020-06-21 | 3 | 5 | +------------+--------------+-------------+--------------+-------------+Items
表: +------------+----------------+---------------+ | item_id | item_name | item_category | +------------+----------------+---------------+ | 1 | LC Alg. Book | Book | | 2 | LC DB. Book | Book | | 3 | LC SmarthPhone | Phone | | 4 | LC Phone 2020 | Phone | | 5 | LC SmartGlass | Glasses | | 6 | LC T-Shirt XL | T-Shirt | +------------+----------------+---------------+ 输出: +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+ | Category | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday | +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+ | Book | 20 | 5 | 0 | 0 | 10 | 0 | 0 | | Glasses | 0 | 0 | 0 | 0 | 5 | 0 | 0 | | Phone | 0 | 0 | 5 | 1 | 0 | 0 | 10 | | T-Shirt | 0 | 0 | 0 | 0 | 0 | 0 | 0 | +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+ 解释: 在周一(2020-06-01, 2020-06-08),Book分类(ids: 1, 2)下,总共销售了20个单位(10 + 10) 在周二(2020-06-02),Book分类(ids: 1, 2)下,总共销售了5个单位 在周三(2020-06-03),Phone分类(ids: 3, 4)下,总共销售了5个单位 在周四(2020-06-04),Phone分类(ids: 3, 4)下,总共销售了1个单位 在周五(2020-06-05),Book分类(ids: 1, 2)下,总共销售了10个单位,Glasses分类(ids: 5)下,总共销售了5个单位 在周六, 没有商品销售 在周天(2020-06-14, 2020-06-21),Phone分类(ids: 3, 4)下,总共销售了10个单位(5 + 5) 没有销售 T-Shirt 类别的商品
原站题解
mysql 解法, 执行用时: 367 ms, 内存消耗: 0 B, 提交时间: 2023-10-15 23:02:25
# Write your MySQL query statement below 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