力扣-市场分析

大家好，我是空空star，本篇带大家了解一道简单的力扣sql练习题。

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前言

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一、题目：1158. 市场分析

Table: Users

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
此表主键是 user_id。
表中描述了购物网站的用户信息，用户可以在此网站上进行商品买卖。

Table: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| buyer_id      | int     |
| seller_id     | int     |
+---------------+---------+
此表主键是 order_id。
外键是 item_id 和（buyer_id，seller_id）。

Table: Items

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
此表主键是 item_id。

请写出一条SQL语句以查询每个用户的注册日期和在 2019 年作为买家的订单总数。

以 任意顺序 返回结果表。

查询结果格式如下。

示例 1:

输入：
Users 表:
+---------+------------+----------------+
| user_id | join_date  | favorite_brand |
+---------+------------+----------------+
| 1       | 2018-01-01 | Lenovo         |
| 2       | 2018-02-09 | Samsung        |
| 3       | 2018-01-19 | LG             |
| 4       | 2018-05-21 | HP             |
+---------+------------+----------------+
Orders 表:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1        | 2019-08-01 | 4       | 1        | 2         |
| 2        | 2018-08-02 | 2       | 1        | 3         |
| 3        | 2019-08-03 | 3       | 2        | 3         |
| 4        | 2018-08-04 | 1       | 4        | 2         |
| 5        | 2018-08-04 | 1       | 3        | 4         |
| 6        | 2019-08-05 | 2       | 2        | 4         |
+----------+------------+---------+----------+-----------+
Items 表:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+
输出：
+-----------+------------+----------------+
| buyer_id  | join_date  | orders_in_2019 |
+-----------+------------+----------------+
| 1         | 2018-01-01 | 1              |
| 2         | 2018-02-09 | 2              |
| 3         | 2018-01-19 | 0              |
| 4         | 2018-05-21 | 0              |
+-----------+------------+----------------+

二、解题

1.错误示范①

提交SQL

select u1.user_id buyer_id,
u1.join_date,
count(1) orders_in_2019
from Users u1
left join Orders u2 
on u1.user_id=u2.buyer_id and substr(u2.order_date,1,4)='2019'
group by u1.user_id,u1.join_date

运行结果

2.正确示范①

提交SQL

select u1.user_id buyer_id,
u1.join_date,
count(u2.order_id) orders_in_2019
from Users u1
left join Orders u2 
on u1.user_id=u2.buyer_id and substr(u2.order_date,1,4)='2019'
group by u1.user_id,u1.join_date

运行结果

3.错误示范②

提交SQL

select u1.user_id buyer_id,
u1.join_date,
u2.num orders_in_2019
from Users u1
left join (select buyer_id,count(1) numfrom Orders where substr(order_date,1,4)='2019'group by buyer_id
) u2
on u1.user_id=u2.buyer_id

运行结果

4.正确示范②

提交SQL

select u1.user_id buyer_id,
u1.join_date,
ifnull(u2.num,0) orders_in_2019
from Users u1
left join (select buyer_id,count(1) numfrom Orders where substr(order_date,1,4)='2019'group by buyer_id
) u2
on u1.user_id=u2.buyer_id

或者

select u1.user_id buyer_id,
u1.join_date,
# ifnull(u2.num,0) orders_in_2019
case when u2.num is null then 0 else u2.num end as orders_in_2019
from Users u1
left join (select buyer_id,count(1) numfrom Orders where substr(order_date,1,4)='2019'group by buyer_id
) u2
on u1.user_id=u2.buyer_id

或者

select u1.user_id buyer_id,
u1.join_date,
ifnull(u2.num,0) orders_in_2019
from Users u1
left join (select buyer_id,count(1) numfrom Orders where year(order_date)='2019'group by buyer_id
) u2
on u1.user_id=u2.buyer_id

运行结果

5.其他

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总结

错误示范①错在返回的无订单数是仍是1，应该把count(1)改为count(u2.order_id)；
错误示范②错在返回的无订单数是null，应该把null转换成0；
知识点：
取2019年可以用substr(order_date,1,4)=‘2019’，也可以用year(order_date)=‘2019’；
将null转换成0可以用ifnull(u2.num,0)，也可以用case when u2.num is null then 0 else u2.num end；
count(1)和count(指定字段)区别：
count(1) 会统计表中的所有的记录数，同count(*) ，包含字段为null 的记录。
count(指定字段) 会统计该字段在表中出现的次数，忽略字段为null 的情况。不统计字段为null 的记录。