Question

我将传感器读数存储在一个表中，数据是仪表读数：

CaptureDate               SensorID      Value
2020-01-11 14:15:33.350   121           23908,0000
2020-01-11 14:00:33.300   123           23161,0000
2020-01-11 14:00:33.240   121           23901,0000
2020-01-11 13:45:33.137   123           23154,0000
2020-01-11 13:45:33.073   121           23894,0000
2020-01-11 13:30:32.927   123           23147,0000

我需要使用SQL获取通过SensorID过滤的一个月的每日总计，以获取类似于以下内容的信息：

Date        SensorID    Value
2020-01-10  121         319
2020-01-11  121         249
2020-01-12  121         289
2020-01-13  121         263
2020-01-14  121         314
2020-01-15  121         248

我尝试获取按天分组的最小值和最大值，但无法获得计数器的差值以获得净值；

SELECT * 
FROM Records
WHERE CaptureDate in 
(
    SELECT min(CaptureDate)
    FROM Records
    WHERE SensorID = 124
        AND convert(date, CaptureDate) >= '2020-01-01'
    GROUP BY convert(date, CaptureDate)
) OR CaptureDate IN (
    SELECT Max(CaptureDate)
    FROM Records
    WHERE SensorID = 124
        AND convert(date, CaptureDate) >= '2020-01-01'
    GROUP BY convert(date, CaptureDate)
) ORDER BY CaptureDate

并返回：

CaptureDate               SensorID  Value   
2020-01-08 14:20:39.627   121       23601.0000
2020-01-08 17:50:39.843   121       23678.0000
2020-01-09 08:50:19.473   121       23678.0000
2020-01-09 18:05:20.300   121       23707.0000
2020-01-10 08:46:06.903   121       23707.0000
2020-01-10 18:15:20.007   121       23796.0000

Answer 1

尝试

 SELECT TimeStamp, SUM(tot) as Tot
 FROM MyTable 
 GROUP BY TimeStamp

Answer 2

我还没有测试过，但是解决方案应该与此类似。您创建了两个不同的表，一个表的每日最小值，另一个表的每日最大值，然后将它们加入：

SELECT mx.CaptureDate, mx.value - mn.value FROM
(
    SELECT CaptureDate, min(CaptureDate) value
    FROM Records
    WHERE SensorID = 124
        AND convert(date, CaptureDate) >= '2020-01-01'
    GROUP BY convert(date, CaptureDate)
) mn, (
    SELECT CaptureDate, Max(CaptureDate) value
    FROM Records
    WHERE SensorID = 124
        AND convert(date, CaptureDate) >= '2020-01-01'
    GROUP BY convert(date, CaptureDate)
) mx 
WHERE mx.CaptureDate = mn.CaptureDate
ORDER BY mx.CaptureDate

Answer 3

我找到了解决方案！没有什么比暴露问题澄清我的想法更好了

SELECT SensorID, CAST(CaptureDate AS DATE) AS Date, MAX(Value)-MIN(Value) As dValue
FROM Readings
WHERE SensorID = 121
GROUP BY CAST(CaptureDate AS DATE), SensorID
ORDER BY CaptureDate

现在多么容易！：D

Answer 4

这应该有效：

演示： DB Fiddle

select 
  convert(date, capturedate) date_only, 
  sensorid, 
  min(value) min_val, 
  max(value) max_val,
  max(value) - min(value) diff
from records
group by convert(date, capturedate), sensorid

Answer 5

我建议尝试将汇总计数存储在单独的列中，然后在该列上运行查询。我不认为“汇总依据”分组会持续存在并传递给外部查询。

获取每天的最小值和最大值之间的差

5 个答案: