[英]Calculating average minutes for items within different time frames
我有一個數據集,需要將其分為兩個子集。 每個子集只能包含特定時間范圍內的項目。 然后,我需要取MAX(上次編輯時間)-MIN(項目添加時間)並除以子集中的項目數量。 因此,我的目標是計算兩個時間范圍內處理所有項目所需的平均時間。 看到圖片了!
我試過了,但是似乎不起作用-結果不正確。 該查詢將在WHERE子句中概述的整個時間段內生成MIN和MAX的結果(因此,幾天的時間會極大地破壞結果)。
SELECT
CASE
WHEN TO_CHAR(ADDED, 'HH24') BETWEEN 09 AND 11
THEN TRUNC(((MAX(MODIFIED) - MIN(ADDED))*24*60)/COUNT(TRANSACTIONS))
ELSE 0
END
+
CASE
WHEN TO_CHAR(ADDED, 'HH24') BETWEEN 15 AND 19
THEN TRUNC(((MAX(MODIFIED) - MIN(ADDED))*24*60)/COUNT(TRANSACTIONS))
ELSE 0
END AS subsets_average
FROM TABLE
我知道你在做什么。 case
陳述書放在錯誤的位置。 你想聚合函數中的條件:
SELECT TRUNC(((MAX(MODIFIED END)-
MIN(ADDED ))*24*60)/COUNT(TRANSACTIONS) as grand_average,
TRUNC(((MAX(TO_CHAR(ADDED, 'HH24') BETWEEN 09 AND 11 THEN MODIFIED END)-
MIN(TO_CHAR(ADDED, 'HH24') BETWEEN 09 AND 11 THEN ADDED END))*24*60)/COUNT(TRANSACTIONS)
) +
TRUNC(((MAX(TO_CHAR(ADDED, 'HH24') BETWEEN 15 AND 19 THEN MODIFIED END)-
MIN(TO_CHAR(ADDED, 'HH24') BETWEEN 15 AND 19 THEN ADDED END))*24*60)/COUNT(TRANSACTIONS) AS subsets_average
FROM TABLE
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.