MySQL高阶1843-可疑银行账户

目录

题目

准备数据

​分析数据

实现

总结

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题目

如果一个账户在 连续两个及以上 月份的 总收入 超过最大收入（max_income），那么认为这个账户 可疑。  账户当月 总收入 是当月存入资金总数（即 transactions 表中 type 字段的 'Creditor'）。

编写一个解决方案，报告所有的 可疑 账户。

准备数据

Create table If Not Exists Accounts (account_id int, max_income int);
Create table If Not Exists Transactions (transaction_id int, account_id int, type ENUM('creditor', 'debtor'), amount int, day datetime);
Truncate table Accounts;
insert into Accounts (account_id, max_income) values ('3', '21000');
insert into Accounts (account_id, max_income) values ('4', '10400');
Truncate table Transactions;
insert into Transactions (transaction_id, account_id, type, amount, day) values ('2', '3', 'Creditor', '107100', '2021-06-02 11:38:14');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('4', '4', 'Creditor', '10400', '2021-06-20 12:39:18');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('11', '4', 'Debtor', '58800', '2021-07-23 12:41:55');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('1', '4', 'Creditor', '49300', '2021-05-03 16:11:04');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('15', '3', 'Debtor', '75500', '2021-05-23 14:40:20');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('10', '3', 'Creditor', '102100', '2021-06-15 10:37:16');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('14', '4', 'Creditor', '56300', '2021-07-21 12:12:25');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('19', '4', 'Debtor', '101100', '2021-05-09 15:21:49');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('8', '3', 'Creditor', '64900', '2021-07-26 15:09:56');
insert into Transactions (transaction_id, account_id, type, amount, day) values ('7', '3', 'Creditor', '90900', '2021-06-14 11:23:07');

accounts表

transactions表

 分析数据

第一步:将日期的月提取出来,并且过滤出type = 'creditor'

select
    account_id,type,amount,
        substr(day,1,4) year,
       substr(day,6,2) month
       from Transactions
where type = 'creditor'
order by account_id,month;

第二步:将两张表进行关联,根据month,t1.account_id,max_income分组

with t1 as (
    select
        account_id,type,amount,
        substr(day,1,4) year,
        substr(day,6,2) month
    from Transactions
    where type = 'creditor'
    order by account_id,month
),t2 as (
    select
        t1.account_id,sum(amount) max,month,max_income
    from t1,Accounts
    where t1.account_id = Accounts.account_id
    group by month,t1.account_id,max_income
    order by t1.account_id)
select * from t2;

第三步:利用开窗函数row_number()形成一列排名

with t1 as (
    select
        account_id,type,amount,
        substr(day,1,4) year,
        substr(day,6,2) month
    from Transactions
    where type = 'creditor'
    order by account_id,month
),t2 as (
    select
        t1.account_id,sum(amount) max,month,max_income
    from t1,Accounts
    where t1.account_id = Accounts.account_id
    group by month,t1.account_id,max_income
    order by t1.account_id
),t3 as (
    select t2.account_id,t2.max,month,
           row_number() over(partition by t2.account_id order by t2.month) rn1
    from t2
    where max > max_income)select * from t3;

第四步:再得出一个等差数列month2,若是差值相同,则连续

with t1 as (
    select
        account_id,type,amount,
        substr(day,1,4) year,
        substr(day,6,2) month
    from Transactions
    where type = 'creditor'
    order by account_id,month
),t2 as (
    select
        t1.account_id,sum(amount) max,month,max_income
    from t1,Accounts
    where t1.account_id = Accounts.account_id
    group by month,t1.account_id,max_income
    order by t1.account_id
),t3 as (
    select t2.account_id,t2.max,month,
           row_number() over(partition by t2.account_id order by t2.month) rn1
    from t2
    where max > max_income
),t4 as (
    select account_id,(month - rn1) month2,
           count(month-rn1) as r2
    from t3
    group by account_id,(month - rn1)
)select * from t4;

实现

with t1 as (
    select
        account_id,type,amount,
        substr(day,1,4) year,
        substr(day,6,2) month
    from Transactions
    where type = 'creditor'
    order by account_id,month
),t2 as (
    select
        t1.account_id,sum(amount) max,month,max_income
    from t1,Accounts
    where t1.account_id = Accounts.account_id
    group by month,t1.account_id,max_income
    order by t1.account_id
),t3 as (
    select t2.account_id,t2.max,month,
           row_number() over(partition by t2.account_id order by t2.month) rn1
    from t2
    where max > max_income
),t4 as (
    select account_id,(month - rn1) month2,
           count(month-rn1) as r2
           from t3
    group by account_id,(month - rn1)
)select
     distinct account_id
     from t4
where r2 >=2;

总结

遇见连续性问题，需要两列差值相同的，最后进行相减，相同的即为连续。