Mysql 使用GROUP BY时包括缺少的（零计数）行

我有一个接收短信的应用程序。我想做的是用mysql做一个统计，统计一小时内的邮件数。例如，早上7点我收到了10条短信，早上8点我收到了20条等等。我的表有以下列ID、smsText、smsDate。。。（其他并不重要）。运行此脚本时：

SELECT HOUR(smsDate), COUNT(ID) FROM SMS_MESSAGES GROUP BY HOUR(smsDate)

它显示了我每小时收到多少信息。问题是，当我没有收到任何消息时，例如在下午5点，此语句不会返回计数为0的第17行，我的结果如下：

Hour Count
...
15 10
16 5
18 2
...

 SELECT DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR, COUNT(ID)
   FROM SMS_MESSAGES 
   JOIN ( 
        SELECT mintime + INTERVAL seq.seq HOUR AS msghour
          FROM (
                SELECT MIN(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS mintime,
                       MAX(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS maxtime
                  FROM SMS_MESSAGES
               ) AS minmax
          JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(HOUR,mintime,maxtime)
       ) AS sq 
     ON DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR = msghour
  GROUP BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR
  ORDER BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR

DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
  FROM seq_0_to_9 a
  JOIN seq_0_to_9 b
  JOIN seq_0_to_9 c
);

DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
  FROM seq_0_to_999 a
  JOIN seq_0_to_999 b
);

，我想要的是这个

Hour Count
...
15 10
16 5
17 0
18 2
...

---

我在网上搜索了一个解决方案，是关于UNION的，但我不知道如何在我的网站上实现它。希望有人能帮助我。

您可以创建一个包含所有小时的表并加入这些表：

CREATE TABLE IF NOT EXISTS `hours` (
  `hour` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `hours` (`hour`) VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9), (10), (11), (12), (13), (14), (15), (16), (17), (18), (19), (20), (21), (22), (23);

SELECT hours.hour, count( SMS_MESSAGES.ID ) 
FROM hours
LEFT JOIN SMS_MESSAGES ON ( hours.hour = HOUR( SMS_MESSAGES.smsDate ) ) 
GROUP BY 1 

---

正如hellocode回答的那样，创建一个包含小时值的新表是一个好方法，下面是通过使用union实现这一点的另一种方法

select t.`hour`,count(s.ID) from (
select 0 as `hour`
union
select 1 as `hour`
union
select 2 as `hour`
union
.
.
.
select 23 as `hour`
) t
left join SMS_MESSAGES s on(t.`hour` = hour(s.smsDate))
group by t.`hour`

观察：

HOUR（）

仅从时间戳中提取小时。查询中可能需要日期和时间。这个答案提供了日期和时间

您需要一种方法来获取一个虚拟表，其中包含适当范围内的所有每小时时间戳。然后需要将该表连接到聚合查询

第一件事：这里有一个查询，它将获得范围内的时间戳

SELECT mintime + INTERVAL seq.seq HOUR AS msghour
  FROM (
        SELECT MIN(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS mintime,
               MAX(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS maxtime
          FROM SMS_MESSAGES
       ) AS minmax
  JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(HOUR,mintime,maxtime)

我们可以获取具有连续运行的每小时时间戳的表

然后我们将其加入到您的查询中。这就是它开始看起来更复杂的地方。我们正在这样做，大致如下：

 SELECT DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR, COUNT(ID)
   FROM SMS_MESSAGES 
   JOIN ( /*the query above wit the sequence of timestamps*/) AS sq 
     ON DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR = msghour
  GROUP BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR
  ORDER BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR

综上所述，它看起来是这样的：

Hour Count
...
15 10
16 5
18 2
...

 SELECT DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR, COUNT(ID)
   FROM SMS_MESSAGES 
   JOIN ( 
        SELECT mintime + INTERVAL seq.seq HOUR AS msghour
          FROM (
                SELECT MIN(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS mintime,
                       MAX(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS maxtime
                  FROM SMS_MESSAGES
               ) AS minmax
          JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(HOUR,mintime,maxtime)
       ) AS sq 
     ON DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR = msghour
  GROUP BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR
  ORDER BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR

DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
  FROM seq_0_to_9 a
  JOIN seq_0_to_9 b
  JOIN seq_0_to_9 c
);

DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
  FROM seq_0_to_999 a
  JOIN seq_0_to_999 b
);

然后，我们可以创建一个视图，将该表与其自身连接起来，以生成1000个组合，如下所示：

Hour Count
...
15 10
16 5
18 2
...

 SELECT DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR, COUNT(ID)
   FROM SMS_MESSAGES 
   JOIN ( 
        SELECT mintime + INTERVAL seq.seq HOUR AS msghour
          FROM (
                SELECT MIN(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS mintime,
                       MAX(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS maxtime
                  FROM SMS_MESSAGES
               ) AS minmax
          JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(HOUR,mintime,maxtime)
       ) AS sq 
     ON DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR = msghour
  GROUP BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR
  ORDER BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR

DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
  FROM seq_0_to_9 a
  JOIN seq_0_to_9 b
  JOIN seq_0_to_9 c
);

DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
  FROM seq_0_to_999 a
  JOIN seq_0_to_999 b
);

最后，我们可以将包含1000个数字的表与它自身连接起来，创建一个视图，该视图将生成一百万个这样的组合：

Hour Count
...
15 10
16 5
18 2
...

 SELECT DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR, COUNT(ID)
   FROM SMS_MESSAGES 
   JOIN ( 
        SELECT mintime + INTERVAL seq.seq HOUR AS msghour
          FROM (
                SELECT MIN(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS mintime,
                       MAX(DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR) AS maxtime
                  FROM SMS_MESSAGES
               ) AS minmax
          JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(HOUR,mintime,maxtime)
       ) AS sq 
     ON DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR = msghour
  GROUP BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR
  ORDER BY DATE(smsDate) + INTERVAL HOUR(smsDate) HOUR

DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
  FROM seq_0_to_9 a
  JOIN seq_0_to_9 b
  JOIN seq_0_to_9 c
);

DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
  FROM seq_0_to_999 a
  JOIN seq_0_to_999 b
);

这里有一封信提供了更多关于这一切的信息