[英]SQL SUM hours between two dates & GROUP BY Column Name?
我需要显示一个月内和前一个月内某项操作经过的总小时数,如下所示:
___________________________________________
| Rank | Action | Month | Prev Month |
|------|----------|------------|------------|
| 1 | Action1 | 580.2 | 200.7 |
| 2 | Action8 | 412.5 | 550.2 |
| 3 | Action10 | 405.0 | 18.1 |
---------------------------------------------
我有一个SQL表,其格式为:
_____________________________________________________
| Action | StartTime | EndTime |
|---------|---------------------|---------------------|
| Action1 | 2015-02-03 06:01:53 | 2015-02-03 06:12:05 |
| Action1 | 2015-02-03 06:22:16 | 2015-02-03 06:25:33 |
| Action2 | 2015-02-03 06:36:07 | 2015-02-03 06:36:49 |
| Action1 | 2015-02-03 06:36:46 | 2015-02-03 06:48:10 |
| ..etc | 20..-..-.. ...etc | 20..-..-.. ...etc |
-------------------------------------------------------
查询是什么样的?
编辑:
ツ的答案使我朝着正确的方向前进,但是我使用JOIN解决了问题。 请参阅下面的解决方案。
我改变了一些值,因为只有一天很无聊
INSERT INTO yourtable
([Action], [StartTime], [EndTime])
VALUES
('Action1', '2015-02-18 06:01:53', '2015-02-18 06:12:05'),
('Action1', '2015-02-18 06:22:16', '2015-02-18 06:25:33'),
('Action2', '2015-04-03 06:36:07', '2015-04-03 06:36:49'),
('Action1', '2015-03-19 06:36:46', '2015-03-19 06:48:10'),
('Action2', '2015-04-13 06:36:46', '2015-04-13 06:48:10'),
('Action2', '2015-04-14 06:36:46', '2015-04-14 06:48:10')
;
现在定义日期边框:
declare @dateEntry datetime = '2015-04-03';
declare @date1 date
, @date2 date
, @date3 date;
set @date1 = @dateEntry; -- 2015-04-03
set @date2 = dateadd(month,-1,@date1); -- 2015-03-03
set @date3 = dateadd(month,-1,@date2); -- 2015-02-03
所选日期将包含所有在2015-04-03 00:00之前开始并在2015-02-03 00:00之后开始的操作
select date1 = @date1
, date2 = @date2
, date3 = @date3
, [Action]
, thisMonth =
sum(
case when Starttime between @date2 and @date1
then datediff(second, starttime, endtime)/360.0
end)
, lastMonth =
sum(
case when Starttime between @date3 and @date2
then datediff(second, starttime, endtime)/360.0
end)
from yourtable
where starttime between @date3 and @date1
group by [Action]
在研究了有关SQL Server的更多信息之后,我只是在重新讨论这个问题。
可以从一个查询创建一个临时表,然后在另一个查询(如果需要的话,一个嵌套查询)中使用它。
这样,结果可以像通过任何其他普通表一样JOIN
在一起,而无需讨厌的CASE
语句。 这对于显示第一个查询所需的其他数据COUNT(DISTINCT ColumnName)
例如COUNT(DISTINCT ColumnName)
也很有用
SELECT TOP 10
t1.Action, t1.[Time],
COALESCE(t2.[Time],0) AS [Previous Period 'Time'],
COALESCE( ( ((t1.[Time]*1.0) - (t2.[Time]*1.0)) / (t2.[Time]*1.0) ), -1 ) AS [Change]
FROM
(
SELECT
Action,
SUM(DATEDIFF(SECOND, StartTime, EndTime)) AS [Time],
FROM Actions
WHERE StartTime BETWEEN @start AND @end
GROUP BY Action
) t1
LEFT JOIN
(
SELECT
Action,
SUM(DATEDIFF(SECOND, StartTime, EndTime)) AS [Time]
FROM Actions
WHERE StartTime BETWEEN @prev AND @start
GROUP BY Action
) t2
ON
t1.Action = t2.Action
ORDER BY t1.[Time] DESC
希望这些信息对某人有所帮助。
可能您应该检查对预处理数据的使用分组,如下所示:
select Action, SUM(Hours)
from (select Action, DATEDIFF('hh',StartTime, EndTime) as Hours
FROM Actions)
group by Action
我的假设是您开始-结束时间跨度太短,您不必担心跨度2个月的日期,因此您可能需要这样的东西:
select
dense_rank() over (order by Month desc) as Rank,
action,
Month,
PrevMonth
from
(
select
action,
sum(case when StartTime >= @curMonth then hours else 0 end) as Month,
sum(case when StartTime >= @prevMonth and StartTime < @curMonth then hours else 0 end) as PrevMonth
from
(
select
action,
StartTime,
datediff(second, StartTime, EndTime) / 3600.0 as hours
from
yourtable
) T1
group by
action
) T2
这将计算持续时间为秒,然后将其除以3600得到小时。 排名仅基于当前月份。 期望您有两个变量@curMonth和@prevMonth,它们具有限制的日期,并且没有将来的数据。
用于测试的SQL Fiddle: http ://sqlfiddle.com/#!6/d64b7d/1
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