Question

我有下表：

SELECT * FROM tblTemp;
+-----+---------------------+
| id  | timestamp           |
+-----+---------------------+
|  84 | 2013-07-18 22:36:32 |
|  85 | 2013-07-18 22:36:33 |
|  88 | 2013-07-18 22:36:33 |
|  89 | 2013-07-18 22:36:34 |
| 263 | 2013-08-04 16:01:05 |
| 264 | 2013-08-04 16:01:11 |
| 272 | 2013-08-04 16:01:42 |
| 273 | 2013-08-04 16:01:42 |
| 274 | 2013-08-04 16:01:42 |
| 275 | 2013-08-04 16:21:25 |
| 276 | 2013-08-04 16:21:25 |
| 279 | 2013-08-04 16:21:26 |
| 280 | 2013-08-04 16:21:27 |
| 281 | 2013-08-04 16:21:27 |
+-----+---------------------+
14 rows in set (0.00 sec)

首先，我想删除视图中的所有双时间戳，如下所示

SELECT DISTINCT timestamp FROM tblTemp;
+---------------------+
| timestamp           |
+---------------------+
| 2013-07-18 22:36:32 |
| 2013-07-18 22:36:33 |
| 2013-07-18 22:36:34 |
| 2013-08-04 16:01:05 |
| 2013-08-04 16:01:11 |
| 2013-08-04 16:01:42 |
| 2013-08-04 16:21:25 |
| 2013-08-04 16:21:26 |
| 2013-08-04 16:21:27 |
+---------------------+
9 rows in set (0.00 sec)

但这还不够。我需要计数，每小时数据库中有多少条目（没有双打） - 关于这个：

+---------------------+-------+
| timestamp           | count |
+---------------------+-------+
| 2013-07-18 22:00:00 | 3     |
| 2013-08-04 16:00:00 | 6     |
+---------------------+-------|

首先我想我可以使用以下命令：

SELECT timestamp, COUNT(id) FROM tblTemp GROUP BY YEAR(timestamp), MONTH(timestamp), DAY(timestamp), HOUR(timestamp)

但它不能正常工作。该命令计算所有行：/

+---------------------+-------+
| timestamp           | count |
+---------------------+-------+
| 2013-07-18 22:36:32 |     4 |
| 2013-08-04 16:01:05 |    10 |
+---------------------+-------+

如果列时间戳中的值为2013-07-18 22:00:00或2013-07-18 22:36:32或其他任何内容，只要时间和日期可识别，则不重要。< / p>

谢谢：）

Answer 1

SELECT DATE_FORMAT(timestamp,'%m/%d/%Y %H') AS date_hour, 
  COUNT(DISTINCT timestamp) 
FROM tblTemp 
GROUP BY date_hour

请参阅http://dev.mysql.com/doc/refman/5.6/en/group-by-functions.html#function_count-distinct

Answer 2

可能是这样的（您可能需要检查日期格式字符串）

SELECT DATE_FORMAT(timestamp,'%m/%d/%Y %H') timestamp, COUNT(id) 
FROM tblTemp 
GROUP BY DATE_FORMAT(timestamp,'%m/%d/%Y %H')

Answer 3

select DATE_FORMAT(`timestamp`, '%Y-%m-%d %H'), count(*)
 from (select distinct `timestamp` from Table1) abc
group by DATE_FORMAT(`timestamp`, '%Y-%m-%d %H');

<强> fiddle

MySQL - 查询 - 在DISTINCT和GROUP BY之间混合

3 个答案: