mySql：统计一列有相同数据的行数
[英]mySql: count number of rows that have the same data in a column
问题描述
我正在尝试 select 表中的所有内容，并计算表中具有相同数据的行数。
SELECT *, COUNT(thedate) daycount FROM `table` ORDER BY thedate DESC
我希望有一个查询输出日期和与该日期关联的行数，循环的 output 将是这样的：
2000 年 1 月 1 日（2 行）
col1, col2, col3, col4
col1, col2, col3, col42000 年 1 月 1 日（3 行）
col1, col2, col3, col4
col1, col2, col3, col4
col1, col2, col3, col42000 年 1 月 1 日（6 行）
col1, col2, col3, col4
col1, col2, col3, col4
col1, col2, col3, col4
col1, col2, col3, col4
col1, col2, col3, col4
col1, col2, col3, col4
ETC...
这有意义吗？
5 个解决方案
解决方案1
3 已采纳 2011-07-01 19:49:49
如果您有一个如下所示的表：
CREATE TABLE yourtable
(
datefield DATETIME,
col1 VARCHAR(20),
col2 INT NOT NULL,
col3 TINYINT NOT NULL,
col4 CHAR(5)
);
并且您想要每天重复 col1.. col4 的计数，您将运行此查询
SELECT
COUNT(datefield) datefield_count,
LEFT(all_fields,10) datefield,
SUBSTR(all_fields,11) all_other_fields
FROM
(
SELECT
DATE(datefield) datefield,
CONCAT(DATE(datefield),'|',
COALESCE(col1,'< NULL >'),'|',
COALESCE(col2,'< NULL >'),'|',
COALESCE(col3,'< NULL >'),'|',
COALESCE(col4,'< NULL >'),'|') all_fields
FROM
yourtable
) A
GROUP BY all_fields;
以下是一些示例数据和查询结果：
mysql> DROP TABLE IF EXISTS yourtable;
Query OK, 0 rows affected (0.04 sec)
mysql> CREATE TABLE yourtable
-> (
-> datefield DATETIME,
-> col1 VARCHAR(20),
-> col2 INT,
-> col3 TINYINT,
-> col4 CHAR(5)
-> );
Query OK, 0 rows affected (0.11 sec)
mysql> INSERT INTO yourtable VALUES
-> (DATE(NOW() - INTERVAL 1 DAY),'rolando',4,3 ,'angel'),
-> (DATE(NOW() - INTERVAL 1 DAY),'rolando',4,3 ,'angel'),
-> (DATE(NOW() - INTERVAL 1 DAY),'rolando',4,3 ,'angel'),
-> (DATE(NOW() - INTERVAL 1 DAY),'rolando',4,NULL,'angel'),
-> (DATE(NOW() - INTERVAL 1 DAY),'rolando',4,NULL,'angel'),
-> (DATE(NOW() - INTERVAL 2 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 2 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 2 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 2 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 2 DAY),'rolando',4,NULL,'edwards'),
-> (DATE(NOW() - INTERVAL 2 DAY),'rolando',4,NULL,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',5,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',5,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'pamela' ,4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'pamela' ,4,NULL,'edwards'),
-> (DATE(NOW() - INTERVAL 3 DAY),'pamela' ,5,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'pamela' ,5,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',4,2 ,'angel'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',4,NULL,'edwards'),
-> (DATE(NOW() - INTERVAL 3 DAY),'rolando',4,NULL,'angel')
-> ;
Query OK, 22 rows affected, 3 warnings (0.03 sec)
Records: 22 Duplicates: 0 Warnings: 3
mysql> SELECT * FROM yourtable;
+---------------------+---------+------+------+-------+
| datefield | col1 | col2 | col3 | col4 |
+---------------------+---------+------+------+-------+
| 2011-06-30 00:00:00 | rolando | 4 | 3 | angel |
| 2011-06-30 00:00:00 | rolando | 4 | 3 | angel |
| 2011-06-30 00:00:00 | rolando | 4 | 3 | angel |
| 2011-06-30 00:00:00 | rolando | 4 | NULL | angel |
| 2011-06-30 00:00:00 | rolando | 4 | NULL | angel |
| 2011-06-29 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-29 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-29 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-29 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-29 00:00:00 | rolando | 4 | NULL | edwar |
| 2011-06-29 00:00:00 | rolando | 4 | NULL | angel |
| 2011-06-28 00:00:00 | rolando | 5 | 2 | angel |
| 2011-06-28 00:00:00 | rolando | 5 | 2 | angel |
| 2011-06-28 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-28 00:00:00 | pamela | 4 | 2 | angel |
| 2011-06-28 00:00:00 | pamela | 4 | NULL | edwar |
| 2011-06-28 00:00:00 | pamela | 5 | 2 | angel |
| 2011-06-28 00:00:00 | pamela | 5 | 2 | angel |
| 2011-06-28 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-28 00:00:00 | rolando | 4 | 2 | angel |
| 2011-06-28 00:00:00 | rolando | 4 | NULL | edwar |
| 2011-06-28 00:00:00 | rolando | 4 | NULL | angel |
+---------------------+---------+------+------+-------+
22 rows in set (0.00 sec)
mysql> SELECT
-> COUNT(datefield) datefield_count,
-> LEFT(all_fields,10) datefield,
-> SUBSTR(all_fields,11) all_other_fields
-> FROM
-> (
-> SELECT
-> DATE(datefield) datefield,
-> CONCAT(DATE(datefield),'|',
-> COALESCE(col1,'< NULL >'),'|',
-> COALESCE(col2,'< NULL >'),'|',
-> COALESCE(col3,'< NULL >'),'|',
-> COALESCE(col4,'< NULL >'),'|') all_fields
-> FROM
-> yourtable
-> ) A
-> GROUP BY all_fields;
+-----------------+------------+----------------------------+
| datefield_count | datefield | all_other_fields |
+-----------------+------------+----------------------------+
| 1 | 2011-06-28 | |pamela|4|2|angel| |
| 1 | 2011-06-28 | |pamela|4|< NULL >|edwar| |
| 2 | 2011-06-28 | |pamela|5|2|angel| |
| 3 | 2011-06-28 | |rolando|4|2|angel| |
| 1 | 2011-06-28 | |rolando|4|< NULL >|angel| |
| 1 | 2011-06-28 | |rolando|4|< NULL >|edwar| |
| 2 | 2011-06-28 | |rolando|5|2|angel| |
| 4 | 2011-06-29 | |rolando|4|2|angel| |
| 1 | 2011-06-29 | |rolando|4|< NULL >|angel| |
| 1 | 2011-06-29 | |rolando|4|< NULL >|edwar| |
| 3 | 2011-06-30 | |rolando|4|3|angel| |
| 2 | 2011-06-30 | |rolando|4|< NULL >|angel| |
+-----------------+------------+----------------------------+
12 rows in set (0.00 sec)
mysql>
我会把它留给你富有想象力的创造力来循环并打印
- 日期字段
- datefield_count
- 打印 all_other_fields 'datefield_count' 次
试试看 ！！！
解决方案2
1 2013-03-21 18:44:32
这不是OP所要求的，而是我遇到这个问题时一直在寻找的。 也许有些人会发现它很有用。
select *
from thetable
join (
select thedate, count( thedate ) as cnt
from thetable
group by thedate
) as counts
using( thedate )
order by thedate
上面的查询将 select 带有一个额外字段cnt的所有内容，其中包含具有相同日期的记录数。 然后打印如下内容是微不足道的：
某个日期，该日期的 2 条记录
col1, col2, col3, col4
col1, col2, col3, col4
其他日期，该日期的 3 条记录
col1, col2, col3, col4
col1, col2, col3, col4
col1, col2, col3, col4
还有其他日期，该日期的 1 条记录
col1, col2, col3, col4
解决方案3
0 2011-07-01 18:31:38
SELECT ...
FROM yourtable
GROUP BY DATE(datefield)
ORDER BY COUNT(DATE(datefield)) DESC
请注意，我使用的是 DATE() function，以防您的日期字段实际上是日期时间。 如果您按日期时间分组，它将完全按 yyyy-mm-dd hh:mm:ss 分组，而不仅仅是 yyyy-mm-dd，您会得到完全不同的结果。
这将为您提供核心结果。 根据需要执行 output 需要在脚本中进行一些后处理，但并不难。 只需缓冲找到的行，直到日期更改，然后 output 缓冲行数。
解决方案4
0 2011-07-01 18:31:57
SELECT *, COUNT(thedate) daycount
FROM `table`
GROUP BY thedate
ORDER BY thedate DESC
解决方案5
0 2011-07-01 18:32:55
SELECT thedate, COUNT(id)
FROM table
WHERE 1
GROUP BY thedate
ORDER BY thedate
计算在mysql中具有相同值的行，然后显示项目号
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