选择按午夜时间分组的数据

时间:2015-07-07 13:02:14

标签: sql oracle group-by

我有一张表:

ID       TIMEVALUE
-----    -------------
1        06.07.15 06:43:01,000000000
2        06.07.15 12:17:01,000000000
3        06.07.15 18:21:01,000000000
4        06.07.15 23:56:01,000000000
5        07.07.15 04:11:01,000000000
6        07.07.15 10:47:01,000000000
7        07.07.15 12:32:01,000000000
8        07.07.15 14:47:01,000000000

我希望特殊时间对这些数据进行分组 我当前的查询如下所示:

SELECT TO_CHAR(TIMEVALUE, 'YYYY\MM\DD'), COUNT(ID), 
  SUM(CASE WHEN TO_CHAR(TIMEVALUE, 'HH24MI') <=700 THEN 1 ELSE 0 END) as morning,
  SUM(CASE WHEN TO_CHAR(TIMEVALUE, 'HH24MI') >700 AND TO_CHAR(TIMEVALUE, 'HH24MI') <1400 THEN 1 ELSE 0 END) as daytime,
  SUM(CASE WHEN TO_CHAR(TIMEVALUE, 'HH24MI') >=1400 THEN 1 ELSE 0 END) as evening FROM Table
WHERE TIMEVALUE >= to_timestamp('05.07.2015','DD.MM.YYYY')
GROUP BY TO_CHAR(TIMEVALUE, 'YYYY\MM\DD')

我得到了这个输出

day          overall     morning    daytime    evening 
-----        ---------
2015\07\05   454         0          0          454
2015\07\06   599         113        250        236
2015\07\07   404         139        265        0

因此在同一天(0-7点,7-14点和14-24点)进行精细分组 但我现在的问题是: 如何在午夜进行分组?

例如,从第二天的6-14,14-23和23-6算起。

我希望你理解我的问题。如果有更好的解决方案,欢迎您甚至改进我的上层查询。

3 个答案:

答案 0 :(得分:1)

编辑:现在已经过测试:SQL Fiddle

关键是调整group by,以便在早上6点之前的任何内容与前一天分组。在那之后,计数非常简单。

SELECT TO_CHAR(CASE WHEN EXTRACT(HOUR FROM timevalue) < 6
                    THEN timevalue - 1
                    ELSE timevalue
                    END, 'YYYY\MM\DD') AS day, 
       COUNT(*) AS overall, 
       SUM(CASE WHEN EXTRACT(HOUR FROM timevalue) >= 6 AND EXTRACT(HOUR FROM timevalue) < 14
                THEN 1 ELSE 0 END) AS morning,
       SUM(CASE WHEN EXTRACT(HOUR FROM timevalue) >= 14 AND EXTRACT(HOUR FROM timevalue) < 23
                THEN 1 ELSE 0 END) AS daytime,
       SUM(CASE WHEN EXTRACT(HOUR FROM timevalue) < 6 OR EXTRACT(HOUR FROM timevalue) >= 23
                THEN 1 ELSE 0 END) AS evening
FROM my_table
WHERE timevalue >= TO_TIMESTAMP('05.07.2015','DD.MM.YYYY')
GROUP BY TO_CHAR(CASE WHEN EXTRACT(HOUR FROM timevalue) < 6
                    THEN timevalue - 1
                    ELSE timevalue
                    END, 'YYYY\MM\DD');

答案 1 :(得分:1)

从时间值减去1天,首先是低于'06:00'的时间,然后是:

SQLFiddle demo

select TO_CHAR(day, 'YYYY\MM\DD') day, COUNT(ID) cnt, 
    SUM(case when '23' < tvh or  tvh <= '06' THEN 1 ELSE 0 END) as midnight,
    SUM(case when '06' < tvh and tvh <= '14' THEN 1 ELSE 0 END) as daytime,
    SUM(case when '14' < tvh and tvh <= '23' THEN 1 ELSE 0 END) as evening
  FROM (
    select id, to_char(TIMEVALUE, 'HH24') tvh,
        trunc(case when (to_char(timevalue, 'hh24') <= '06') 
                   then timevalue - interval '1' day  
                   else timevalue end) day
      from t1
    )
  GROUP BY day

答案 2 :(得分:0)

也许你可以这样做(有些重新格式化或PIVOT):

WITH spans AS 
    (SELECT TIMESTAMP '2015-01-01 00:00:00' + LEVEL * INTERVAL '1' HOUR AS start_time
    FROM dual
    CONNECT BY TIMESTAMP '2015-01-01 00:00:00' + LEVEL * INTERVAL '1' HOUR < LOCALTIMESTAMP),
t AS 
    (SELECT start_time, lead(start_time, 1) OVER (ORDER BY start_time) AS end_time, ROWNUM AS N
    FROM spans
    WHERE EXTRACT(HOUR FROM start_time) IN (6,14,23))
SELECT N, start_time, end_time, COUNT(*) AS ID_COUNT,
    DECODE(EXTRACT(HOUR FROM start_time), 6,'morning', 14,'daytime', 23,'evening') AS daytime
FROM t
    JOIN YOUR_TABLE WHERE TIMEVALUE BETWEEN start_time AND end_time
GROUP BY N;

当然,初始时间值(我的示例中为“2015-01-01 00:00:00”)必须低于表格中的最小日期。