动态计算同一行的时间差

Question

是否有任何方法可以根据 'DOWN' 和 'UP' 值计算同一列中行之间 SQL 的时间差，如下所示：

有 3 种情况（我知道）：

黄橙绿：有一个state_id2(down)，后面有一个state_id5(up)，所以需要计算两行的时间差；
蓝色：有多个state_id2(down)，后面有一个state_id5(up)，所以需要计算第一行和最后一行的时间差；
红色：只有一个state_id2(down)，因为任何更新都还是down，所以要计算到月底的时差

希望你能帮帮我。

Answer 1

我之前的做法是

Select a.state_time as downtime, 
    (
        select min(inner.state_time) from tablename downentry where
        inner.state_time > outer.state_time and downentry.state='UP'
    ) as uptime

from tablename upentry 
where state = 'DOWN'

然后你需要找到它们之间的datediff，如果uptime为null，则downtime和'endofmonth'

之间的datediff

它的性能可能很差，所以我总是将答案写到数据仓库中，但我认为它会给出您要求的结果。

Answer 2

首先考虑为此使用 LAG。

但是使用累积 SUM，MIN 的 window 版本也适用于超过 2 个 DOWN：

-- test reference data
declare @State table (id int, state varchar(4));
insert into @State (id, state) values 
(2,'DOWN'),
(5,'UP')

-- test data, using a table variable
declare @AlertState table (alert_id int identity(1,1), host_id int, state_time datetime, state_id int);
insert into @AlertState (host_id, state_time, state_id) values 
(119, GetDate()-0.32, 2),
(119, GetDate()-0.31, 5),
(119, GetDate()-0.24, 2),
(119, GetDate()-0.23, 2),
(119, GetDate()-0.22, 2),
(119, GetDate()-0.21, 5),
(119, GetDate()-0.15, 5),
(119, GetDate()-0.11, 2);

-- The query
select alert_id, host_id, state_time, state_id,
 diff_min = (
   case 
   when state_id = 5 then 
     datediff(minute, min(state_time) over (partition by host_id, stategroup), state_time)
   when state_id = 2 and stategroup is null then
     datediff(minute, state_time, cast(EOMONTH(GetDate()) as datetime)+1)
   end),
 s.state
from (
    select alert_id, host_id, state_time, state_id,
    sum(case state_id when 5 then 1 end) over (partition by host_id order by state_time desc) as stategroup
    from @AlertState
    where state_id in (2,5)
) q
left join @State s on s.id = q.state_id
order by state_time, alert_id;

Answer 3

SQL2012+

您可以尝试以下解决方案：

SELECT y.group_id, host_id = MIN(host_id), start_time = MIN(state_time), end_time = MAX(state_time), diff_minute = DATEDIFF(MINUTE, MIN(state_time), MAX(state_time))
FROM (
    SELECT *, group_id = SUM(x.new_group_start) OVER(ORDER BY x.host_id, x.state_time)
    FROM (
        SELECT  *, new_group_start = IIF(a.state_id = 'DOWN' AND ISNULL(LAG(a.state_id) OVER(ORDER BY a.host_id, a.state_time), 'UP') = 'UP', 1, 0)
        FROM    @Alerts a
    ) x
) y
GROUP BY y.group_id
ORDER BY y.group_id

Demo

动态计算同一行的时间差

Calculate the time difference in the same row dynamically

sql

sql-server

datediff

availability