总和每月全天每天每小时的总利用率答案

【问题标题】：sum totals utilization for each hour in day for all day in month总和每月全天每天每小时的总利用率
【发布时间】：2016-02-27 16:16:01
【问题描述】：

我有以下使用 PrestoDB 的 SQL 查询：

SELECT 
  date_trunc('month',TimeFrom) AS Month,
  date_trunc('hour',TimeFrom) AS HourFrom,
  date_trunc('hour',TimeTo) AS HourTo,
  CASE
      WHEN resource = 'c81d9eb67e93' THEN 'S'
      WHEN resource = 'cce1eee4c73f' THEN 'L'
      WHEN resource = 'fbc1d63e0f15' THEN 'G'
      WHEN resource = '5ef75e433db2' THEN 'Q'
      WHEN resource = '4b6ba65cbe14' THEN 'T'
      WHEN resource = 'df0a8b60467d' THEN 'R'
      WHEN resource = '013472b63651' THEN 'C'
      WHEN resource = '7464b77d93be' THEN 'F'
      WHEN resource = '6f1186bdbc12' THEN 'M'
      ELSE 'other resource'
    END AS Resource,
    COUNT (*) AS Utilization
FROM (SELECT 
       resource,
       date_add('hour', i - 1, mytable.TimeFrom) AS TimeFrom, 
       date_add('hour', i, mytable.TimeFrom) AS TimeTo
          FROM (SELECT 1 AS i UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4
                UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8
                UNION ALL SELECT 9 UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
                UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15 UNION ALL SELECT 16
                UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19 UNION ALL SELECT 20
                UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23 UNION ALL SELECT 24
          ) AS numbers
          INNER JOIN (SELECT
            resource,
            date_trunc(
              'hour',
              FROM_UNIXTIME(timefrom)
            ) AS TimeFrom,
            date_trunc(
              'hour',
              (FROM_UNIXTIME(timeto) + interval '45' minute)
            ) AS TimeTo
          FROM 
            reservation
          WHERE
            type = 'create'
          ORDER BY
            date_trunc(
              'day',
              FROM_UNIXTIME(timefrom)
            ),
            resource) AS mytable ON numbers.i <= date_diff('hour', TimeFrom, TimeTo)
          ORDER BY 
            TimeFrom,
            resource)
GROUP BY 
  date_trunc('month',TimeFrom),
  date_trunc('hour',TimeFrom),
  date_trunc('hour',TimeTo),
  resource

它像这样返回表：

    Month | HourFrom | HourTo | Resource | Utilization
    ["2015-08-01 00:00:00.000","2015-08-05 14:00:00.000","2015-08-05 15:00:00.000","S",4]
    ["2015-08-01 00:00:00.000","2015-08-05 17:00:00.000","2015-08-05 18:00:00.000","Q",1]
    ["2015-08-01 00:00:00.000","2015-08-05 17:00:00.000","2015-08-05 18:00:00.000","Q",1]
    ["2015-08-01 00:00:00.000","2015-08-05 18:00:00.000","2015-08-05 19:00:00.000","S",5]
    ["2015-08-01 00:00:00.000","2015-08-05 20:00:00.000","2015-08-05 21:00:00.000","Q",1]
    ["2015-08-01 00:00:00.000","2015-08-05 21:00:00.000","2015-08-05 22:00:00.000","Q",1]
    ["2015-08-01 00:00:00.000","2015-08-05 22:00:00.000","2015-08-05 23:00:00.000","Q",1]
    ["2015-08-01 00:00:00.000","2015-08-06 00:00:00.000","2015-08-06 01:00:00.000","L",2]
    ["2015-08-01 00:00:00.000","2015-08-06 00:00:00.000","2015-08-06 01:00:00.000","Q",2]
    ["2015-08-01 00:00:00.000","2015-08-06 02:00:00.000","2015-08-06 03:00:00.000","S",3]

如您所见，HourFrom 和 HourTo 列按天分开。有没有办法聚合它们，以便我得到一个月内所有日子每小时的总和/总计数？

假设我有 1 月和 2 月的数据，我们只在凌晨 1 点到凌晨 4 点开放，并且只有 2 个资源，那么我想要获得的决赛桌看起来像这样（即 Jan1AM - Jan2AM 给了我利用率number[total] 在 1 月的那个时间范围内的所有天数连续）：

Month | HourFrom | HourTo | Resource | Utilization
Jan     Jan 1AM    Jan 2AM     S         number
Jan     Jan 1AM    Jan 2AM     X         number
Jan     Jan 2AM    Jan 3AM     S         number
Jan     Jan 2AM    Jan 3AM     X         number
Jan     Jan 3AM    Jan 4AM     S         number
Jan     Jan 3AM    Jan 4AM     X         number
Feb     Feb 1AM    Feb 2AM     S         number
Feb     Feb 1AM    Feb 2AM     X         number
Feb     Feb 2AM    Feb 3AM     S         number
Feb     Feb 2AM    Feb 3AM     X         number
Feb     Feb 3AM    Feb 4AM     S         number
Feb     Feb 3AM    Feb 4AM     X         number

如何更改我的原始查询以执行此类操作？

【问题讨论】：

标签： sql database relational-database presto

【解决方案1】：

我将从一个更简单的查询开始，该查询使用 24 小时制的小时数，并且不会费心填写缺失的小时数。这将是：

SELECT date_trunc('month', TimeFrom) AS Month,
       hour(TimeFrom) as HourFrom,
       (CASE WHEN resource = 'c81d9eb67e93' THEN 'S'
             WHEN resource = 'cce1eee4c73f' THEN 'L'
             WHEN resource = 'fbc1d63e0f15' THEN 'G'
             WHEN resource = '5ef75e433db2' THEN 'Q'
             WHEN resource = '4b6ba65cbe14' THEN 'T'
             WHEN resource = 'df0a8b60467d' THEN 'R'
             WHEN resource = '013472b63651' THEN 'C'
             WHEN resource = '7464b77d93be' THEN 'F'
             WHEN resource = '6f1186bdbc12' THEN 'M'
             ELSE 'other resource'
        END) AS Resource,
       COUNT(*) AS Utilization
FROM reservation r
WHERE type = 'create'
GROUP BY date_trunc('month', TimeFrom), hour(TimeFrom), resource;

也许这个版本对于你想要完成的事情已经足够了。

【讨论】：