计算每个客户和值的连续行数

Question

我关注table：

create table tickets (id int, customer text, status smallint);

insert into tickets values
(1, 'A', 0),
(2, 'A', 0),
(3, 'B', 0),
(4, 'B', 0),
(5, 'A', 0),
(6, 'B', 1),
(7, 'C', 0),
(8, 'B', 0),
(9, 'A', 0),
(10, 'B', 0),
(11, 'A', 1),
(12, 'A', 1),
(13, 'A', 1);

我试图根据这个线程找出我的案例：

但我能做到的最好的是这个，这还不是我的预期输出

select customer, status, count(*) as cnt
from (select t.*,
             (row_number() over (order by id) -
              row_number() over (partition by customer, status order by id)
             ) as grp
      from tickets t
     ) x
group by grp, customer, status
order by max(id)

DBFiddle

预期输出 将是：（每个客户的每个状态的连续记录）

+-----------+-------+-----+
| customer  | status| cnt |
+-----------+-------+-----+
|  A        | 0     |   4 |
|  A        | 1     |   3 |
|  B        | 0     |   2 |
|  B        | 1     |   1 |
|  B        | 0     |   2 |
|  C        | 0     |   1 |

Answer 1

对于差距和孤岛的解决方案，首先row_number您需要按客户进行分区。

SELECT customer,    status, COUNT(*) FROM (
select t.*,
             (row_number() over (partition by customer order by id) -
              row_number() over (partition by customer, status order by id)
             ) as grp
      from tickets t
) X
GROUP BY customer,  status, grp
ORDER BY customer,  max(id)

dbfiddle

结果：

customer    status  count
--------    ------  -----
A           0       4
A           1       3
B           0       2
B           1       1
B           0       2
C           0       1

计算每个客户和值的连续行数

Count consecutive rows for each customer and value

sql

postgresql

row-number