[英]Calculating average minutes for items within different time frames
我有一个数据集,需要将其分为两个子集。 每个子集只能包含特定时间范围内的项目。 然后,我需要取MAX(上次编辑时间)-MIN(项目添加时间)并除以子集中的项目数量。 因此,我的目标是计算两个时间范围内处理所有项目所需的平均时间。 看到图片了!
我试过了,但是似乎不起作用-结果不正确。 该查询将在WHERE子句中概述的整个时间段内生成MIN和MAX的结果(因此,几天的时间会极大地破坏结果)。
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
陈述书放在错误的位置。 你想聚合函数中的条件:
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
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