SQL 组中两行之间的差异

Question

我有一个商店id、处理批次id和开始时间的表records如下：

|store_id | batch_id | process_start_time |
| A       | 1        | 10                 |
| B       | 1        | 40                 |
| C       | 1        | 30                 |
| A       | 2        | 400                |
| B       | 2        | 800                |
| C       | 2        | 600                |
| A       | 3        | 10                 |
| B       | 3        | 80                 |
| C       | 3        | 90                 |

这里，需要按batch_id和time_taken分组的行是store A和store C的process_start_time的差异。

所以，预期的结果是：

batch_id | time_taken
1        | 20
2        | 200
3        | 80

我试图做类似的事情：

select batch_id, ((select process_start_time from records where store_id = 'C') - (select process_start_time from records where store_id = 'A')) as time_taken 
from records group by batch_id;

但无法弄清楚该特定组中的 select 特定行。

感谢您的关注！

更新：对于商店 C，process_start_time 列不一定是最大值

Answer 1

您似乎需要条件聚合和算术：

select batch_id,
       (max(case when store_id = 'C' then process_start_time end) -
        min(case when store_id = 'A' then process_start_time end)
       ) as diff
from records
group by batch_id;

Answer 2

您可以尝试自我加入。

SELECT r1.batch_id,
       r1.process_start_time - r2.process_start_time time_taken
       FROM records r1
            INNER JOIN records r2
                       ON r1.batch_id = r2.batch_id
       WHERE r1.store_id = 'C'
             AND r2.store_id = 'A';

Answer 3

这是另一个答案。 这是使用记录表的两个实例，我们将它们与 where 子句链接起来，如下所示：

select a.batch_id,
       c.process_start_time - a.process_start_time as time_taken
from   records a,
       records c
where  a.store_id = 'A'
and    c.store_id = 'C'
and    exists (
select 1
from   records x
where  x.batch_id = a.batch_id
and    x.batch_id = c.batch_id
);

Answer 4

SELECT DISTINCT 
    store_a.batch_id,
    store_c.process_start_time - store_a.process_start_time AS 'time_taken'
    FROM records store_a  
    INNER JOIN records store_c
    ON store_a.batch_id = store_c.batch_id 
       AND store_c.store_id = 'C' 
       AND  store_a.store_id = 'A'