Question

我使用PrestoDB进行以下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

它返回如下表：

    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]

正如您所看到的，HourFrom和HourTo列按天划分。有没有办法聚合它们，以便我得到一个月内所有日子的每小时总数/总数？

所以说我有1月和1月的数据。 2月份，我们只在凌晨1点到4点开放，只有2个资源，然后我想得到的决赛桌看起来像这样（即Jan1AM - Jan2AM给出了我在该时间段内所有日子的利用率[总计]在一月一行）：

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

如何更改原始查询以执行此类操作？

Answer 1

我将从一个更简单的查询开始，该查询使用24小时制作时间，并且无需填写缺少的小时数。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;

也许这个版本足以满足您的要求。

总计每月全天每小时的利用率

1 个答案: