DateTime按日期和小时分组

I have a datetime called activity_dt and the data looks like this:

我有一个名为activity_dt的日期时间，数据如下所示：

2/5/2013 9:24:00 AM
2/7/2013 7:17:00 AM

How do i group by date and hour?

我如何按日期和小时分组？

3 个解决方案

#1

SQL Server :

SQL Server：

SELECT [activity_dt], count(*)
FROM table1
GROUP BY DATEPART(day, [activity_dt]), DATEPART(hour, [activity_dt]);

Oracle :

甲骨文：

SELECT [activity_dt], count(*)
FROM table1
GROUP BY TO_CHAR(activity_dt, 'DD'), TO_CHAR(activity_dt, 'hh');

MySQL :

MySQL：

SELECT [activity_dt], count(*)
FROM table1
GROUP BY hour( activity_dt ) , day( activity_dt )

#2

Using MySQL I usually do it that way:

使用MySQL我通常这样做：

SELECT count( id ), ...
FROM quote_data
GROUP BY date_format( your_date_column, '%Y%m%d%H' )
order by your_date_column desc;

Or in the same idea, if you need to output the date/hour:

或者在同一个想法中，如果您需要输出日期/小时：

SELECT count( id ) , date_format( your_date_column, '%Y-%m-%d %H' ) as my_date
FROM  your_table 
GROUP BY my_date
order by your_date_column desc;

If you specify an index on your date column, MySQL should be able to use it to speed up things a little.

如果在日期列上指定索引，MySQL应该能够使用它来加速一些事情。

#3

SELECT [activity_dt], COUNT(*) as [Count]
  FROM 
 (SELECT dateadd(hh, datediff(hh, '20010101', [activity_dt]), '20010101') as [activity_dt]
    FROM table) abc
 GROUP BY [activity_dt]

#1