计算项目在不同时间范围内的平均分钟数

Question

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我有一个数据集，需要分成两个子集。每个子集必须仅包括特定时间范围内的项。然后，我需要取最大(上次编辑的时间)-MIN(项目添加时间)，并除以子集中的项目数量。因此，我的目标是计算在两个时间框架内处理所有项目所需的平均时间。看这张照片！

我尝试过这样做，但结果似乎不正确；查询在WHERE子句中为MIN和MAX生成整个时间段的结果(因此，多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

Answer 1

我明白你想做什么了。case语句在错误的位置。您需要聚合函数中的条件：

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