LeetCode:Database 115.仓库经理
Posted Xiao Miao
tags:
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要求:写一个 SQL 查询来报告, 每个仓库的存货量是多少立方英尺。
表: Warehouse的结构
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| name | varchar |
| product_id | int |
| units | int |
+--------------+---------+
(name, product_id) 是该表主键.
该表的行包含了每个仓库的所有商品信息.
表: Products的结构
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| product_id | int |
| product_name | varchar |
| Width | int |
| Length | int |
| Height | int |
+---------------+---------+
product_id 是该表主键.
该表的行包含了每件商品以英尺为单位的尺寸(宽度, 长度和高度)信息.
Warehouse 表:
+------------+--------------+-------------+
| name | product_id | units |
+------------+--------------+-------------+
| LCHouse1 | 1 | 1 |
| LCHouse1 | 2 | 10 |
| LCHouse1 | 3 | 5 |
| LCHouse2 | 1 | 2 |
| LCHouse2 | 2 | 2 |
| LCHouse3 | 4 | 1 |
+------------+--------------+-------------+
Products 表:
+------------+--------------+------------+----------+-----------+
| product_id | product_name | Width | Length | Height |
+------------+--------------+------------+----------+-----------+
| 1 | LC-TV | 5 | 50 | 40 |
| 2 | LC-KeyChain | 5 | 5 | 5 |
| 3 | LC-Phone | 2 | 10 | 10 |
| 4 | LC-T-Shirt | 4 | 10 | 20 |
+------------+--------------+------------+----------+-----------+
Result Table:
+----------------+------------+
| WAREHOUSE_NAME | VOLUME |
+----------------+------------+
| LCHouse1 | 12250 |
| LCHouse2 | 20250 |
| LCHouse3 | 800 |
+----------------+------------+
Id为1的商品(LC-TV)的存货量为 5x50x40 = 10000
Id为2的商品(LC-KeyChain)的存货量为 5x5x5 = 125
Id为3的商品(LC-Phone)的存货量为 2x10x10 = 200
Id为4的商品(LC-T-Shirt)的存货量为 4x10x20 = 800
仓库LCHouse1: 1个单位的LC-TV + 10个单位的LC-KeyChain + 5个单位的LC-Phone.
总存货量为: 1*10000 + 10*125 + 5*200 = 12250 立方英尺
仓库LCHouse2: 2个单位的LC-TV + 2个单位的LC-KeyChain.
总存货量为: 2*10000 + 2*125 = 20250 立方英尺
仓库LCHouse3: 1个单位的LC-T-Shirt.
总存货量为: 1*800 = 800 立方英尺.
SQL语句:
with a as(
select product_id,width*length*height as
area
from products
)
select b.name as WAREHOUSE_NAME,sum(b.units*a.area) as VOLUME
from a
join warehouse b
on a.product_id=b.product_id
group by b.name;
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