Question

我有一个带有“日期”列的表。我想按小时分组到特定日期。

Answer 1

你也可以这样做：

SELECT TRUNC(datecol, 'HH24') FROM mytable
GROUP BY TRUNC(datecol, 'HH24');

Answer 2

Select TO_CHAR(date,'HH24')
from table
where date = TO_DATE('20110224', 'YYYYMMDD')
group by TO_CHAR(date,'HH24')

Answer 3

select to_char(datecol,'HH24') thehour, count(*) count_in_hour
from tbl
where datecol = date '20110224'
group by to_char(datecol,'HH24')
order by thehour asc

Answer 4

或者，我会建议EXTRACT (datetime)，但是（empahsis mine）：

如果请求HOUR，MINUTE或SECOND，则expr必须评估为表达式数据类型TIMESTAMP，TIMESTAMP WITH TIME ZONE，TIMESTAMP 与当地时区，或间隔日到第二。 DATE无效这里，因为Oracle数据库将其视为ANSI DATE数据类型没时间了。

因此，结果有点难看：

  SELECT EXTRACT (HOUR FROM CAST (m.a_date AS TIMESTAMP)) AS hour_,
         COUNT (*) AS count_
    FROM MY_TABLE m
GROUP BY EXTRACT (HOUR FROM CAST (m.a_date AS TIMESTAMP));

按日期（小时）的SQL组

4 个答案: