SQL 区间内的聚合和平均值

Question

我有一个 table 具有这样的数据集

Time_1	Temperature
2020-11-20 12:05:15.520	75
2020-11-20 12:06:50.297	111
2020-11-20 12:08:10.513	147
2020-11-20 12:11:10.593	72
2020-11-20 12:10:46.000	62
2020-11-20 12:15:15.263	171
2020-11-20 12:16:34.577	238
2020-11-20 12:16:45.577	284
2020-11-20 12:18:08.373	304
2020-11-20 12:20:57.030	299
2020-11-20 12:22:21.670	313
2020-11-20 12:34:43.733	100

我想根据另一个时间戳(time_2)的时间间隔(大多数情况下是10分钟间隔)对上述温度数据进行平均，得到类似这样的东西

Time_2	Records_in_interval	Av_Temp
2020-11-20 12:05:19.000	6	106.33
2020-11-20 12:15:19.000	5	287.6
2020-11-20 12:25:19.000	1	100

我面临的一个问题是间隔并不总是 10 分钟。这是我到目前为止所做的，但它不起作用。

DECLARE @StartDate nvarchar(20)
DECLARE @EndDate nvarchar(20)
     
    SET @StartDate ='2020-11-20 12:00:00'
    SET @EndDate =  '2020-11-20 23:59:59'

SELECT  
      [Date1]= CONVERT(datetime, t2.[time_2], 20),
      DATEADD(MINUTE, DATEDIFF(MINUTE,  t2.[time_2], t1.[time_1]) / 10 * 10,t2.[time_2]) AS [date_truncated],
      COUNT(*) AS [records_in_interval],
      AVG(t1.[temperature]) AS [Av_Temp] 

FROM fix.dbo.main_db2 t2
    LEFT JOIN fix1.dbo.main_db1 t1 on DATEADD(MINUTE, DATEDIFF(MINUTE, t2.[time_2], t1.[time_1]) / 10 * 10,t2.[time_2])=t2.[time_2]
    
    WHERE 
     CONVERT(datetime, t2.[time_2], 20)  BETWEEN CONVERT(datetime, @StartDate , 20) AND CONVERT(datetime, @EndDate , 20)
    
GROUP BY 
    DATEADD(MINUTE, DATEDIFF(MINUTE, t2.[time_2], t1.[time_1]) / 10 * 10, t2.[time_2]),
    t2.[time_2]
ORDER BY 
[date_truncated]

Answer 1

如果你有第二个 table 时间，那么你可以使用 join 和聚合：

select t2.time_2, count(temperature), avg(temperature)
from (select t2.*, lead(time_2) over (order by time_2) as next_time_2
      from table2 t2
     ) t2 left join
     table1 t1
     on t1.time_1 >= t2.time_2 and
        t1.time_1 < t2.next_time_2
group by t2.time_2;

编辑：

我认为您可以为此使用相关子查询：

select t2.time_2, count(temperature), avg(temperature)
from (select t2.*,
             (select min(tt2.time_2)
              from table2 tt2
              where tt2.time_2 > t2.time_2
             ) as next_time_2
      from table2 t2
     ) t2 left join
     table1 t1
     on t1.time_1 >= t2.time_2 and
        t1.time_1 < t2.next_time_2
group by t2.time_2;

但老实说，我不记得 SQL Server 2000 中的限制是什么。在这一点上它完全过时了。

Answer 2

如果您只有第一个 table，您可以使用 union 来获得所有三个结果。

SELECT '2020-11-20 12:05:19.000' AS Time_2,
COUNT(*) AS Records_in_Interval,
AVG(temp) AS Avg_Temp
FROM #temp
WHERE time > '2020-11-20 12:05:19.000' AND time < '2020-11-20 12:15:19.000'
UNION ALL
SELECT '2020-11-20 12:15:19.000' AS Time_2,
COUNT(*) AS Records_in_Interval,
AVG(temp) AS Avg_Temp
FROM #temp
WHERE time > '2020-11-20 12:15:19.000' AND time < '2020-11-20 12:25:19.000'
UNION ALL 
SELECT '2020-11-20 12:25:19.000' AS Time_2,
COUNT(*) AS Records_in_Interval,
AVG(temp) AS Avg_Temp
FROM #temp
WHERE time > '2020-11-20 12:25:19.000'

谢谢

SQL 区间内的聚合和平均值

Aggregation and average value with interval in SQL

sql

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sql-server

average