孤岛和缺口订购问题MYSQL 8.0

我试图使用partition by&row_number（）来计算给定日期范围内的连续重复值。本质上，它试图捕获“条纹”，如果条纹中出现中断，则当值再次出现时，计数应重新开始

复制这些结果的代码如下： 这是电流输出 下面是所需的输出

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我在这件事上绞尽脑汁已经有一段时间了。任何帮助都将不胜感激

您的数据与您的结果不符，因此很难实现您的结果

每日|回复|简短|姓名|行数| :--------- | :------------------ | ------: 2020-09-20 | C | 1 2020-09-21 | A | 1 2020-09-22 | B | 1 2020-09-23 | C | 1 2020-09-24 | C | 2 2020-09-25 | A | 1 2020-09-26 | A | 2 2020-09-27 | A | 3 2020-09-28 | A | 4 dbfiddle

您可以执行以下操作：

select *,
  row_number() over(partition by grp order by daily) as seqnum  
from (
  select *,
    sum(inc) over(order by daily) as grp
  from (
    select *,
      case when lag(response_short_name) over(order by daily) = response_short_name
        then 0 else 1 end as inc
    from partion_test
    order by daily
  ) x
) y
order by daily

结果:

 daily       response_short_name  inc  grp  seqnum 
 ----------- -------------------- ---- ---- ------ 
 2020-09-20  C                    1    1    1      
 2020-09-21  A                    1    2    1      
 2020-09-22  B                    1    3    1      
 2020-09-23  C                    1    4    1      
 2020-09-24  C                    0    4    2      
 2020-09-25  A                    1    5    1      
 2020-09-26  A                    0    5    2      
 2020-09-27  A                    0    5    3      
 2020-09-28  A                    0    5    4      

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请参阅运行示例：

请参阅我认为您只需要按A.daily排序。您的基本问题是子查询没有IRDER，因为表本质上是未排序的，所以请将ORDER BY添加到subquery@nbk将订单添加到子查询仍然会导致“seqnum”无序有时订单是“优化”的，因此使用额外的限制10000

 +------------+---------------------+--------+--+
|   daily    | response_short_name | seqnum |  |
+------------+---------------------+--------+--+
| 2020-09-20 | C                   |      1 |  |
| 2020-09-21 | A                   |      1 |  |
| 2020-09-22 | B                   |      1 |  |
| 2020-09-23 | C                   |      1 |  |
| 2020-09-24 | C                   |      2 |  |
| 2020-09-25 | A                   |      1 |  |
| 2020-09-26 | A                   |      2 |  |
| 2020-09-27 | A                   |      3 |  |
| 2020-09-28 | A                   |      4 |  |
+------------+---------------------+--------+--+

    CREATE TABLE partion_test (
    daily DATE,
    response_short_name VARCHAR(10)
    
);

INSERT INTO `partion_test` (`daily`, `response_short_name`) VALUES
('2020-09-21', 'A'),
('2020-09-25', 'A'),
('2020-09-26', 'A'),
('2020-09-27', 'A'),
('2020-09-28', 'A'),
('2020-09-22', 'B'),
('2020-09-20', 'C'),
('2020-09-23', 'C'),
('2020-09-24', 'C');

select `daily`,`response_short_name`,
       row_number() over (partition by `response_short_name`, grp order by `daily`) as row_num
from (select t.*,
             (row_number() over (order by `daily`) -
              row_number() over (partition by `response_short_name` order by `daily`)
             ) as grp
      from partion_test t
     ) t
     ORDER BY `daily`

daily | response_short_name | row_num :--------- | :------------------ | ------: 2020-09-20 | C | 1 2020-09-21 | A | 1 2020-09-22 | B | 1 2020-09-23 | C | 1 2020-09-24 | C | 2 2020-09-25 | A | 1 2020-09-26 | A | 2 2020-09-27 | A | 3 2020-09-28 | A | 4

select *,
  row_number() over(partition by grp order by daily) as seqnum  
from (
  select *,
    sum(inc) over(order by daily) as grp
  from (
    select *,
      case when lag(response_short_name) over(order by daily) = response_short_name
        then 0 else 1 end as inc
    from partion_test
    order by daily
  ) x
) y
order by daily

 daily       response_short_name  inc  grp  seqnum 
 ----------- -------------------- ---- ---- ------ 
 2020-09-20  C                    1    1    1      
 2020-09-21  A                    1    2    1      
 2020-09-22  B                    1    3    1      
 2020-09-23  C                    1    4    1      
 2020-09-24  C                    0    4    2      
 2020-09-25  A                    1    5    1      
 2020-09-26  A                    0    5    2      
 2020-09-27  A                    0    5    3      
 2020-09-28  A                    0    5    4