题目：报告系统状态的连续日期

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。

编写一个 SQL 查询 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。

Failed table	Succeeded table

最后答案要求：

题目理解：

题目意思不太好理解，数据是2018-12-28到2019-01-06 每天都有一个任务失败或者成功。一般情况下的记录方法是“日期+成功/失败”。

题中是将日期根据任务状态分别放到两个表中：成功的记录到Succeeded table里，失败的记录到Failed table。

解法一：窗口函数分组

解题思路：
1）使用UNION all 将两个表连接在一起，
2）用窗口函数和subdate()来找到分组指标diff。SUBDATE()函数从日期中减去时间/日期间隔，然后返回日期。
3）使用group by 根据type和diff再次分组，用min和max找出起始日期。

select type as period_state, min(date) as start_date, max(date) as end_date
from
(
    select type, date, subdate(date,row_number()over(partition by type order by date)) as diff
    from
    (
        select 'failed' as type, fail_date as date from Failed
        union all
        select 'succeeded' as type, success_date as date from Succeeded
    ) a
)a
where date between '2019-01-01' and '2019-12-31'
group by type,diff
order by start_date

解法二：变量方法

解题思路：
1）两表聚合
2）通过变量给连续状态分组
3）返回需求

# Write your MySQL query statement below
--3、返回需求
SELECT period_state, MIN(date) AS start_date, MAX(date) AS end_date
FROM
--2、通过变量给连续状态分组
    (SELECT date, period_state, IF(@pre_state=period_state, @id, @id:=@id+1) AS id,            @pre_state:=period_state
    FROM
--1、两表聚合
        (SELECT fail_date AS date, 'failed' AS period_state
        FROM Failed
        UNION ALL
        SELECT success_date AS date, 'succeeded' AS period_state
        FROM Succeeded) A, (SELECT @pre_state:=NULL, @id:=0) B
    WHERE date BETWEEN '2019-01-01' AND '2019-12-31'
    ORDER BY date) C
GROUP BY id

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/report-contiguous-dates
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