Oracle中的递归迭代

Question

我有一个这样的表：

+----+-----+------+
| id | ord | test |
+----+-----+------+
|  1 |   1 | A    |
|  1 |   2 | B    |
|  1 |   3 | C    |
|  2 |   1 | B    |
|  2 |   2 | C    |
+----+-----+------+

(这是一些创建数据的代码(

drop table temp_test;
create table temp_test (id varchar(20), ord varchar(20), test varchar(20));
insert into temp_test (id,ord,test) values ('1','1','A');
insert into temp_test (id,ord,test) values ('1','2','B');
insert into temp_test (id,ord,test) values ('1','3','C');
insert into temp_test (id,ord,test) values ('2','1','B');
insert into temp_test (id,ord,test) values ('2','2','C');
commit;

我怎样才能得到以下结果？

+----+-----+-------+
| id | ord | test  |
+----+-----+-------+
|  1 |   1 | A     |
|  1 |   2 | A_B   |
|  1 |   3 | A_B_C |
|  2 |   1 | B     |
|  2 |   2 | B_C   |
+----+-----+-------+

我尝试过使用LAG((，类似于：

select CONCAT(lag(TEST) over (partition by ID order by ord),TEST) AS TEST from temp_test;

但是它不能递归地工作。

此代码有效：

SELECT 
R1.*,
(   SELECT  LISTAGG(test, ';') WITHIN GROUP (ORDER BY ord)
FROM    temp_test R2
WHERE   R1.ord >= R2.ord
AND     R1.ID = R2.ID
GROUP BY ID
) AS WTR_KEYWORD_1
FROM temp_test R1
ORDER BY id, ord;

但对于更大的数据集来说，它的性能还不够。

Answer 1

有些人说层次查询已经过时，但它们通常比递归CTE 执行得更好

SELECT id,
ord,
LTRIM(sys_connect_by_path(test,'_'),'_') as test
FROM temp_test r2 START WITH ord = 1 -- use MIN() to get this if it's not always 1
CONNECT BY PRIOR id = id AND ord = PRIOR ord + 1;

演示

Answer 2

您可以使用递归cte来实现此

with cte(id,ord,test,concat_val) 
as (select id,ord,test,test as concat_val
from temp_test
where ord=1
union all
select a.id,a.ord,a.test,b.concat_val||'_'||a.test
from temp_test a
join cte b
on a.id=b.id
and a.ord=b.ord+1 
)
select * from cte order by id,ord

在这里演示

https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=78baa20f7f364e653899caf63ce7ada2