Question

我们说，我有一张包含服务中断列表的表格字段是：service_id，from_time，to_time。时间是时间戳类型。

具有这些间隔的服务的状态是＆＃34; DOWN＆＃34;。其他时间是＆＃34; UP＆＃34;。

我正在寻找一个查询，该查询将返回一个连续的时间间隔列表，表示给定日期和现在之间的特定服务健康状况。

E.g。该表只包含服务srv1的一次中断：

srv1, 11/01/2017 13:43:32, 11/01/2017 15:20:12, DOWN

然后，来自年度开始的查询结果应如下所示：

srv1, 11/01/2017 15:20:12, 24/07/2017 23:55:00, UP  
srv1, 11/01/2017 13:43:31, 11/01/2017 15:20:12, DOWN  
srv1, 01/01/2017 00:00:00, 11/01/2017 13:43:31, UP

假设中断不重叠后续间隔的from_time等于服务的前一时间间隔的to_time 如果服务当前为DOWN，则中断表中的to_time等于NULL。

查询适用于PG 9.6

Answer 1

基本上，你想要：

选择所有记录
选择“中间”记录，添加相反状态
添加上一个状态

以下使用union all来实现此目的：

select t.*
from t
union all
select id, coalesce(prev_to_time, '2017-01-01'::timestamp) as start_time,
       start_time as to_time,
       (case when status = 'DOWN' then 'UP' else 'DOWN' end) as status
from (select t.*,
             lag(to_time) over (partition by id order by start_time) as prev_to_time,
             lag(status) over (partition by id order by start_time) as prev_status
      from t
     ) t
where (prev_to_time is null and start_time <> '2017-01-01'::timestamp or
       prev_to_time <> start_time
      ) and
      (prev_status is distinct from status)
union all
select id, max(end_time), now(), 'UP'
from t
group by id
having max(end_time) <> now()
order by id, start_time;

Answer 2

您可以使用lag（）函数回顾上一行。

在这种情况下，例如：

with down as (select *, lag(to_time) over (rows unbounded preceding)
                        as last_up
                from outage order by from_time asc),
      full_log as (select host, from_time, to_time, 'down' 
                          AS as status 
                     FROM down
                UNION ALL
                   select host, last_up, from_time, 'up' as status
                     from down)
select * from full_log order by from_time asc;

如何在SELECT查询中填补时间间隔？

2 个答案: