我有一张像:
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
我想按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
这样就可以在同一天分组(0-7 点钟、7-14 点钟和 14-24 点钟)
但我现在的问题是:
我怎样才能在午夜分组?
例如从第二天的 6-14 点、14-23 点和 23-6 点开始计数。
我希望你明白我的问题。如果有更好的解决方案,欢迎您甚至改进我的上层查询。
最佳答案
编辑 : 现已测试:SQL Fiddle
关键只是调整group by
这样早上 6 点之前的任何内容都会与前一天分组。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');
关于sql - 选择按时间分组的午夜数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31269273/