leetcode 数据库练习 - 1205 每月交易I和II

每月交易（一）

Table: Transactions

问题：编写一个 sql 查询来查找每个月和每个国家/地区的事务数及其总金额、已批准的事务数及其总金额。

查询结果格式如下所示：

Transactions table:
+------+---------+----------+--------+------------+
| id | country | state | amount | trans_date |
+------+---------+----------+--------+------------+
| 121 | US | approved | 1000 | 2018-12-18 |
| 122 | US | declined | 2000 | 2018-12-19 |
| 123 | US | approved | 2000 | 2019-01-01 |
| 124 | DE | approved | 2000 | 2019-01-07 |
+------+---------+----------+--------+------------+

Result table:
+----------+---------+-------------+----------------+--------------------+-----------------------+
| month | country | trans_count | approved_count | trans_total_amount | approved_total_amount |
+----------+---------+-------------+----------------+--------------------+-----------------------+
| 2018-12 | US | 2 | 1 | 3000 | 1000 |
| 2019-01 | US | 1 | 1 | 2000 | 2000 |
| 2019-01 | DE | 1 | 1 | 2000 | 2000 |
+----------+---------+-------------+----------------+--------------------+-----------------------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/monthly-transactions-i
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

代码一：

select date_format(T.trans_date,'%Y-%m') as month, T.country,
count(*) as trans_total_count,
sum(T.state='approved') as approved_count,  # 注意，这里使用sum()函数，不能使用count()函数
sum(amount) as trans_total_amount,
sum((T.state='approved')*amount) as approved_total_amount
 
from Transactions4 T
 
group by month(T.trans_date),T.country
order by month , country desc;

代码二: 和代码以相比，select 语句后有点不同，使用if 判断

select date_format(T.trans_date,'%Y-%m') as month, T.country,
count(*) as trans_total_count,
sum(if(T.state='approved',1,0)) as approved_count,  # 注意，这里使用sum()函数，不能使用count()函数
sum(amount) as trans_total_amount,
sum(if(T.state='approved',amount,0)) as approved_total_amount
 
from Transactions4 T
 
group by month(T.trans_date),T.country
order by month , country desc;

每月交易（二）

Transactions 记录表

Chargebacks 表

问题：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
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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
order by a.month;

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：

万能网友：

select date_format(a.trans_date,"%Y-%m") month,
a.country,
sum(a.sta="approved") approved_count,   #这几句只能使用sum()函数，不能使用count()函数
sum((a.sta="approved")*a.amount) approved_amount,
sum(a.sta="backs") chargeback_count,
sum((a.sta="backs")*a.amount) chargeback_amount
from
(select *,"approved" as sta   # 添加了一列, 列名sta
from Transactions
where state="approved"
union all
select c.trans_id,t.country,t.state,t.amount,c.trans_date,"backs" as sta   #也添加了一列，要注意：列名要和前面添加的相同，都是sta
from Chargebacks c
left join transactions t
on t.id=c.trans_id
) as a
group by date_format(a.trans_date,"%Y-%m"),a.country
having sum(a.sta="approved")+sum(a.sta="backs")<>0
order by date_format(a.trans_date,"%Y-%m");

代码分析：

中间过程表的格式：

select *,"approved" as sta   # 添加了一列, 列名sta
from Transactions
where state="approved"
union all
select c.trans_id,t.country,t.state,t.amount,c.trans_date,"backs" as sta   #也添加了一列，要注意：列名要和前面添加的相同，都是sta
from Chargebacks c  left join transactions t  on t.id=c.trans_id

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"]