leetcode 数据库练习 - 1205 每月交易II
作者：互联网
Transactions 记录表
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| id | int |
| country | varchar |
| state | enum |
| amount | int |
| trans_date | date |
+----------------+---------+
id 是这个表的主键。
该表包含有关传入事务的信息。
状态列是类型为 [approved（已批准）、declined（已拒绝）] 的枚举。
Chargebacks 表
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| trans_id | int |
| charge_date | date | 注：这个题目给的是charge_date，但它定义表检查答案的时候定义的是trans_date，所以代码中chargebacks表中这一列也定义成了trans_date
+----------------+---------+
退单包含有关放置在事务表中的某些事务的传入退单的基本信息。
trans_id 是 transactions 表的 id 列的外键。
每项退单都对应于之前进行的交易，即使未经批准。
编写一个 SQL 查询，以查找每个月和每个国家/地区的已批准交易的数量及其总金额、退单的数量及其总金额。
注意：在您的查询中，给定月份和国家，忽略所有为零的行。
查询结果格式如下所示：
Transactions 表：
+------+---------+----------+--------+------------+
| id | country | state | amount | trans_date |
+------+---------+----------+--------+------------+
| 101 | US | approved | 1000 | 2019-05-18 |
| 102 | US | declined | 2000 | 2019-05-19 |
| 103 | US | approved | 3000 | 2019-06-10 |
| 104 | US | approved | 4000 | 2019-06-13 |
| 105 | US | approved | 5000 | 2019-06-15 |
+------+---------+----------+--------+------------+
Chargebacks 表：
+------------+------------+
| trans_id | trans_date |
+------------+------------+
| 102 | 2019-05-29 |
| 101 | 2019-06-30 |
| 105 | 2019-09-18 |
+------------+------------+
Result 表：
+----------+---------+----------------+-----------------+-------------------+--------------------+
| month | country | approved_count | approved_amount | chargeback_count | chargeback_amount |
+----------+---------+----------------+-----------------+-------------------+--------------------+
| 2019-05 | US | 1 | 1000 | 1 | 2000 |
| 2019-06 | US | 3 | 12000 | 1 | 1000 |
| 2019-09 | US | 0 | 0 | 1 | 5000 |
+----------+---------+----------------+-----------------+-------------------+--------------------+
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/monthly-transactions-ii
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个人思路：
select distinct a.month, a.country,ifnull(b.approved_count,0) as approved_count, # 加了distinct，是因为最终结果中可能会有重复值 ifnull(b.approved_amount,0) as approved_amount, ifnull(c.chargeback_count,0) as chargeback_count, ifnull(c.chargeback_amount,0) as chargeback_amount from (select date_format(T.trans_date,'%Y-%m')as month, T.country as country from Transactions as T # 如这张表中有两个相同的月份和国家，结果中就会产生重复值，在这里加distinct没有太大用途，因为和下面一张表Union时，还会产生重复值，所以是在最终结果中加入distinct关键字 group by month(T.trans_date),T.country union all select date_format(C.trans_date,'%Y-%m')as month , T.country as country from Chargebacks as C join Transactions as T on C.trans_id=T.id group by month(C.trans_date),T.country) a left join (select date_format(T.trans_date,'%Y-%m') as month ,T.country, count(T.id) as approved_count, sum(T.amount) as approved_amount from Transactions as T where T.state='approved' group by month(T.trans_date) ,T.country) b on a.month=b.month and a.country=b.country left join (select date_format(C.trans_date,'%Y-%m') as month,T.country, count(T.id) as chargeback_count ,sum(T.amount) as chargeback_amount from Transactions as T join Chargebacks as C on T.id=C.trans_id group by month(C.trans_date),T.country) c on a.month=c.month and a.country=c.country where isnull(b.approved_count)+isnull(c.chargeback_count)<2; # 判断非零行，当count为0时，对应的amount一定也是0，所以当approved_count和chargeback_count都为0时，使用isnull（）函数判断后，这两者相加结果为2，小于2说明，其中一个count有值
代码分析：
a表：这张表中的月份和country是最全的，是在这张表的基础上进行left join的
PS：group by 字段后面有2个字段，select语句查询2个字段，不要忘记查询country； 因为b表和c表中的country可能完全都不相同，此时使用left join不能得到正确的结果，所以select语句查询时包含country字段，因此，group by字段后面也要添加country分组
select date_format(T.trans_date,'%Y-%m')as month, T.country as country from Transactions as T group by month(T.trans_date),T.country union all select date_format(C.trans_date,'%Y-%m')as month , T.country as country from Chargebacks as C join Transactions as T on C.trans_id=T.id group by month(C.trans_date),T.country
运行结果：
b表：
select date_format(T.trans_date,'%Y-%m') as month ,T.country, count(T.id) as approved_count, sum(T.amount) as approved_amount from Transactions as T where T.state='approved' group by month(T.trans_date) ,T.country
运行结果：
c表：
select date_format(C.trans_date,'%Y-%m') as month,T.country, count(T.id) as chargeback_count ,sum(T.amount) as chargeback_amount from Transactions as T join Chargebacks as C on T.id=C.trans_id group by month(C.trans_date),T.country
运行结果：
测试用表：
第一张：
"Transactions":
[[100,"CB","declined",4000,"2019-02-04"]
,[101,"BB","approved",7000,"2019-02-17"]
,[102,"CA","declined",6000,"2019-02-26"]
,[103,"AA","declined",7000,"2019-04-01"]]
"Chargebacks":
[[100,"2019-03-29"],
[102,"2019-02-28"]
,[103,"2019-05-09"]]}}
第二张：这一张是说明最后结果会有重复值，所以在最终结果处加distinct关键字
"Transactions"
:[[101,"US","approved",1000,"2019-05-18"]
,[102,"US","declined",2000,"2019-05-19"],
[103,"US","approved",3000,"2019-06-10"],
[104,"US","declined",4000,"2019-06-13"]
,[105,"US","approved",5000,"2019-06-15"]],
"Chargebacks"
:[[102,"2019-05-29"],
[101,"2019-06-30"],
[105,"2019-09-18"]]}}
第三张表： b表和c表中，使用group by子句时，不要忘了country字段也要分组
"Transactions":
[[100,"BB","declined",2000,"2019-02-25"],
[101,"CA","declined",4000,"2019-02-22"],
[102,"BA","declined",2000,"2019-03-17"]
,[103,"AC","declined",6000,"2019-02-12"]
,[104,"AA","approved",3000,"2019-02-20"]
[105,"AB","declined",7000,"2019-03-22"]],
"Chargebacks
:[[100,"2019-04-26"]
[101,"2019-03-09"],
[103,"2019-03-07"],
[105,"2019-06-12"]]}}
第四张表：
"Transactions"
:[[100,"CB","approved",3000,"2019-04-11"]
,[101,"CC","declined",8000,"2019-04-05"],
[102,"BC","approved",2000,"2019-04-16"],
[103,"BC","approved",6000,"2019-02-18"],
[104,"AA","declined",7000,"2019-04-23"]
,[105,"CA","declined",1000,"2019-03-12"],
[106,"AA","declined",3000,"2019-03-24"],
[107,"CB","declined",2000,"2019-03-24"]],"
Chargebacks"
:[[100,"2019-07-05"]
,[101,"2019-04-15"],
[102,"2019-07-09"],
[103,"2019-03-06"]
,[104,"2019-05-08"]
,[105,"2019-05-12"]
,[106,"2019-04-10"]
,[107,"2019-07-01"]
标签：1205,country,approved,II,2019,month,date,trans,leetcode 来源： https://www.cnblogs.com/bravesunforever/p/11745118.html