获取每天的最小值和最大值之间的差异

Question

I store sensor readings in a table, the data is meter readings:

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

I need to use SQL to get the daily totals for a month filtered by SensorID, to get something similar to this:

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

I have tried to obtain minimum and maximum values grouped by day but I can't get the difference of the counter to obtain the net value;

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

And return:

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

Try that

With DailyMinMax AS
(
  SELECT 
    CAST(CaptureDate AS date) date,
    SensorID,
    MIN(Value) minvalue,
    Max(Value) maxvalue
  FROM Records 
  GROUP BY CAST(CaptureDate AS date), SensorID
)
SELECT 
  date,
  SensorID,
  maxvalue-minvalue AS MaxMinDifference 
FROM DailyMinMax 

Answer 2

I have not tested it, but the solution should be something similar to this. you create two different tables, one with the daily minimum and one with the dailt maximum and then join them:

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

I found the solution! There is nothing better than exposing the problem to clarify my ideas

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

How easy it is now! :D

Answer 4

This should work:

Demo: 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

I suggest try and store aggregated count in a separate column and run your query on that column. I don't think group by aggregates persist and passed to outer query.