LeetCode 【困难】数据库-第1225：报告系统状态的连续日期

题目

数据

结果

解答

1.union 合并数据、筛选日期

        select 'succeeded' as 'state',success_date as 'dt'
        from Succeeded where success_date between '2019-01-01' and '2019-12-31'
        union 
        select 'failed'as 'state',fail_date as 'dt'
        from Failed    where fail_date    between '2019-01-01' and '2019-12-31'

2.根据状态排名，该状态对应的日期减去排名，如果相同，则可以认为日期是连续的。subdate函数（date,a)-->date-a

3.最后，挑选出来，最小（start）、最大（end） 日期，然后按照状态分组、按照subdate函数的结果分组。

select state as period_state,   min(dt ) as start_date,    max(dt) as end_date
from
(
    select *,rank()over(partition by state order by  dt ) as rk ,subdate(dt,rank()over(partition by state order by  dt)) as dif
    from 
    (
        select 'succeeded' as 'state',success_date as 'dt'
        from Succeeded where success_date between '2019-01-01' and '2019-12-31'
        union 
        select 'failed'as 'state',fail_date as 'dt'
        from Failed    where fail_date    between '2019-01-01' and '2019-12-31'
    )t1
)t2
group by state,dif
order by dt