Question

在MySQL中，如何对类似的连续记录进行分组，使它们不被其他传送记录分隔

id | type |
1  | 1    |
2  | 1    |
3  | 1    |
4  | 5    |
5  | 1    |
6  | 8    |
7  | 1    |
8  | 3    |
9  | 3    |
10 | 3    |

结果应该是这样的：

id | type |
3  | 1    |
4  | 5    |
5  | 1    |
6  | 8    |
8  | 1    |
10 | 3    |

Answer 1

  DROP TABLE IF EXISTS my_table;

  CREATE TABLE my_table
  (id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
  ,type INT NOT NULL
  ,x INT NOT NULL
  );

  INSERT INTO my_table VALUES
  (1,1,1),
  (2,1,1),
  (3,1,1),
  (4,5,5),
  (5,1,1),
  (6,8,8),
  (7,1,1),
  (8,1,1);

  SELECT MIN(c.id) id 
       , a.type
       , a.x
    FROM my_table a
    LEFT 
    JOIN my_table b 
      ON b.id + 1 = a.id 
     AND b.type = a.type
     AND b.x = a.x
    LEFT 
    JOIN my_table c 
      ON c.id >= a.id
     AND c.type = a.type
     AND c.x = a.x
    LEFT 
    JOIN my_table d 
      ON d.id - 1 = c.id 
     AND d.type = a.type
     AND d.x = a.x
   WHERE b.id IS NULL 
     AND c.id IS NOT NULL
     AND d.id IS NULL
   GROUP 
      BY a.id; 

  +------+------+---+
  |  id  | type | x |
  +------+------+---+
  |    3 |    1 | 1 |
  |    4 |    5 | 5 |
  |    5 |    1 | 1 |
  |    6 |    8 | 8 |
  |    8 |    1 | 1 |
  +------+------+---+

Answer 2

您只需要一个标准组：

select count(*) id, type, `group`
from mytable
group by type, `group`

请注意，您需要分隔保留字＆＃34; group＆＃34;

Answer 3

你的问题很不对劲。我认为以下可能接近你想要的。它通过分配标识连续组的新type列，根据连续的grp值标识组。

select type, max(id) as maxid, count(*) as numingroup
from (select t.*,
             (@grp := if(@type = type, @grp, if(@type := type, @grp + 1, @grp + 1))) as grp
      from my_table t cross join
           (select @grp := 0, @type := -1)x
      order by id
     ) t
group by grp, type;

将类似的连续记录分组，它们没有被其他传送记录分开 - Mysql

3 个答案: