1225. 报告系统状态的连续日期

SQL架构

Table: Failed

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| fail_date    | date    |
+--------------+---------+
该表主键为 fail_date。
该表包含失败任务的天数.

Table: Succeeded

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| success_date | date    |
+--------------+---------+
该表主键为 success_date。
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

查询结果样例如下所示:

Failed table:
+-------------------+
| fail_date         |
+-------------------+
| 2018-12-28        |
| 2018-12-29        |
| 2019-01-04        |
| 2019-01-05        |
+-------------------+

Succeeded table:
+-------------------+
| success_date      |
+-------------------+
| 2018-12-30        |
| 2018-12-31        |
| 2019-01-01        |
| 2019-01-02        |
| 2019-01-03        |
| 2019-01-06        |
+-------------------+


Result table:
+--------------+--------------+--------------+
| period_state | start date   | end date     |
+--------------+--------------+--------------+
| present      | 2019-01-01   | 2019-01-03   |
| missing      | 2019-01-04   | 2019-01-05   |
| present      | 2019-01-06   | 2019-01-06   |
+--------------+--------------+--------------+

结果忽略了 2018 年的记录，因为我们只关心从 2019-01-01 到 2019-12-31 的记录
从 2019-01-01 到 2019-01-03 所有任务成功，系统状态为 "succeeded"。
从 2019-01-04 到 2019-01-05 所有任务失败，系统状态为 "failed"。
从 2019-01-06 到 2019-01-06 所有任务成功，系统状态为 "succeeded"。

这一题题意给的result有点问题，还是按照控制台的用例结果为准 couchpotato613 发布于 1 个月前 109 阅读 MySQL

我是这样想的：
先按照日期递增的顺序把失败的和成功的天数连续查出来，sql如下：

   select date,period_state from (
   select fail_date as date,'failed' as period_state from failed
   where fail_date between '2019-01-01' and '2019-12-31'
   union all
   select success_date as date,'succeeded' as period_state from Succeeded
   where success_date between '2019-01-01' and '2019-12-31'
   ) t1 order by t1.date as

然后按照连续天数的状态给个排名rank,连续的状态相同的rank一样，不一样的rank递增，
sql如下：

select date,period_state,
if(@prev = period_state ,@tmp := @tmp, @tmp := @tmp + 1) as rank,
@prev := period_state as prev from
(select date,period_state from (
	select fail_date as date,'failed' as period_state from failed
	where fail_date between '2019-01-01' and '2019-12-31'
	union all
	select success_date as date,'succeeded' as period_state from Succeeded
	where success_date between '2019-01-01' and '2019-12-31'
	) t1 order by t1.date asc) t2,(select @prev := null,@tmp := 0) as init

得到上述结果之后，再按照rank分组，把每一组的最大值和最小值分别作为start_date和end_date,
完整的sql如下:

select t3.period_state as period_state,min(date) as start_date,
max(date) as end_date from
(select date,period_state,
if(@prev = period_state ,@tmp := @tmp, @tmp := @tmp + 1) as rank,
@prev := period_state as prev
from
(select date,period_state from (
	select fail_date as date,'failed' as period_state from failed
	where fail_date between '2019-01-01' and '2019-12-31'
	union all
	select success_date as date,'succeeded' as period_state from Succeeded
	where success_date between '2019-01-01' and '2019-12-31'
	) t1 order by t1.date asc) t2,(select @prev := null,@tmp := 0) as init) t3
    group by rank

标签：状态,1225,01,state,period,日期,2019,date,select
来源： https://www.cnblogs.com/leeeee/p/11902026.html