SQL，力扣题目1225，报告系统状态的连续日期【窗口函数】

一、力扣链接

二、题目描述

表：Failed

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+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| fail_date    | date    |
+--------------+---------+
该表主键为 fail_date (具有唯一值的列)。
该表包含失败任务的天数.

表： Succeeded

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+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| success_date | date    |
+--------------+---------+
该表主键为 success_date (具有唯一值的列)。
该表包含成功任务的天数.

系统每天运行一个任务。每个任务都独立于先前的任务。任务的状态可以是失败或是成功。

编写解决方案找出 2019-01-01 到 2019-12-31 期间任务连续同状态 period_state 的起止日期（start_date 和 end_date）。即如果任务失败了，就是失败状态的起止日期，如果任务成功了，就是成功状态的起止日期。

最后结果按照起始日期 start_date 排序

三、目标拆解

四、建表语句

sql 复制代码

Create table If Not Exists Failed (fail_date date)
Create table If Not Exists Succeeded (success_date date)
Truncate table Failed
insert into Failed (fail_date) values ('2018-12-28')
insert into Failed (fail_date) values ('2018-12-29')
insert into Failed (fail_date) values ('2019-01-04')
insert into Failed (fail_date) values ('2019-01-05')
Truncate table Succeeded
insert into Succeeded (success_date) values ('2018-12-30')
insert into Succeeded (success_date) values ('2018-12-31')
insert into Succeeded (success_date) values ('2019-01-01')
insert into Succeeded (success_date) values ('2019-01-02')
insert into Succeeded (success_date) values ('2019-01-03')
insert into Succeeded (success_date) values ('2019-01-06')

五、过程分析

1、过滤日期

2、拼接两个状态的数据

3、差值相等的即为连续

4、计算最小/最大值即为起始/结束日期

六、代码实现

sql 复制代码

with t1 as(
select fail_date date,  'failed'  period_state
from Failed
where fail_date between '2019-01-01' and '2019-12-31'
union all
select success_date,  'succeeded'
from Succeeded
where success_date between '2019-01-01' and '2019-12-31'
)
,t2 as (
    select period_state,
           date,
           row_number() over (order by date)        rn1,
           row_number() over (partition by period_state order by date)        rn2,
           row_number() over (order by date) - row_number() over (partition by period_state order by date) diff
-- 用date - rn2求diff，会造成一年被分成12个月的连续区间
    from t1
)
# select * from t2;
select period_state,
       min(date) start_date,
       max(date) end_date
from t2
group by period_state, diff
order by start_date;

七、结果验证

八、小结

1、与1285 为同一类型题目

2、过滤 -> 新增一列 -> 拼接 -> 排名函数 -> 分组 -> 求最小/大值