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SQL問題求助(修正)

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SQL範例
各位不好意思,昨天我有發問而且已經有解答了,但我漏掉了一個地方,所以修正後重新再發問一次

表1是原始資料,每個測項固定六分鐘會有一筆數值及兩個狀態。
表2是這次想求的結果。以小時GroupBy去計算出平均值,每六分鐘一筆資料,所以一小時會是十筆資料,計算平均值的方式是固定用十筆資料來計算,另外這邊需要去計算這十筆資料中哪個狀態是出現最多次的。
表3是計算狀態1 各個狀態的數量,由前三行可以知道狀態A出現3次,B有3次,C有4次,所以表2第一行的Status1就是出現最多次的狀態'C'。
表4則為狀態2的,算法跟表3一樣,所以表2第一行的Status2為'2'。

跟昨天的差別在於多了一欄狀態需要判斷
再次麻煩各位抽空解答,感謝

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1 個回答

1
rogeryao
iT邦超人 8 級 ‧ 2022-06-22 12:17:49
最佳解答
SELECT A.*,B.Status2
FROM (
-- Status1
SELECT DateTimePerHour,MonitorID,SubTotal / Status1CountSum AS 'AverageValue',Status1
FROM (
SELECT DateTimePerHour,MonitorID,Status1,
SUM(Status1Count) OVER (PARTITION BY DateTimePerHour,MonitorID) AS 'Status1CountSum',
SUM(ValuePart) OVER (PARTITION BY DateTimePerHour,MonitorID) AS 'SubTotal',
ROW_NUMBER() OVER (PARTITION BY DateTimePerHour,MonitorID ORDER BY Status1Count DESC) AS 'NO'
FROM (
SELECT dateadd(hour, datediff(hour, 0, [Date_Time]), 0) AS 'DateTimePerHour',
MonitorID,Status1,COUNT(*) AS 'Status1Count',SUM(Value) AS 'ValuePart'
FROM tmp
GROUP BY dateadd(hour, datediff(hour, 0, [Date_Time]), 0),MonitorID,Status1
) AS K
) AS Q
WHERE Q.NO = 1 ) AS A
--
INNER JOIN (
-- Status2
SELECT DateTimePerHour,MonitorID,
-- SubTotal / Status2CountSum AS 'AverageValue',
Status2
FROM (
SELECT DateTimePerHour,MonitorID,Status2,
-- SUM(Status2Count) OVER (PARTITION BY DateTimePerHour,MonitorID) AS 'Status2CountSum',
-- SUM(ValuePart) OVER (PARTITION BY DateTimePerHour,MonitorID) AS 'SubTotal',
ROW_NUMBER() OVER (PARTITION BY DateTimePerHour,MonitorID ORDER BY Status2Count DESC) AS 'NO'
FROM (
SELECT dateadd(hour, datediff(hour, 0, [Date_Time]), 0) AS 'DateTimePerHour',
MonitorID,Status2,COUNT(*) AS 'Status2Count'
-- ,SUM(Value) AS 'ValuePart'
FROM tmp
GROUP BY dateadd(hour, datediff(hour, 0, [Date_Time]), 0),MonitorID,Status2
) AS K
) AS Q
WHERE Q.NO = 1 
) AS B ON B.DateTimePerHour = A.DateTimePerHour AND B.MonitorID = A.MonitorID
-- AND B.AverageValue = A.AverageValue

Demo

CooperWu iT邦新手 3 級 ‧ 2022-06-22 15:26:32 檢舉

謝謝恐龍大神回答,不過我想再等等看有沒有其他答案

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