使用递归自连接“修复”根PK

Question

取以下虚构的网站数据（0表示缺失值）：

USER_ID	TRANSACTION_ID_PARENT	TRANSACTION_ID_CHILD
1	0	1
1	1	2
1	2	3
1	0	11
1	11	12

如何添加“修复”/生成根主键的列：

USER_ID	TRANSACTION_ID_PARENT	TRANSACTION_ID_CHILD	TRANSACTION_ID_ROOT
1	0	1	1
1	1	2	1
1	2	3	1
1	0	11	11
1	11	12	11

我的预感是按 USER_ID 分组的递归 CTE？如何在 IBM db2 方言中实现这一点？

也许递归自连接有更优雅的数据库特定变体？

Answer 1

可以这样做：

with table1 (USER_ID, TRANSACTION_ID_PARENT, TRANSACTION_ID_CHILD) as (
  VALUES
  (1, 0, 1),
  (1, 1, 2),
  (1, 2, 3),
  (1, 0, 11),
  (1, 11, 12)
),
rec (USER_ID, TRANSACTION_ID_PARENT, TRANSACTION_ID_CHILD, TRANSACTION_ID_ROOT) as (
  select USER_ID, TRANSACTION_ID_PARENT, TRANSACTION_ID_CHILD, TRANSACTION_ID_CHILD
  from table1 where TRANSACTION_ID_PARENT= 0
  union all
  select rec.USER_ID, table1.TRANSACTION_ID_PARENT, table1.TRANSACTION_ID_CHILD, rec.TRANSACTION_ID_ROOT
  from rec, table1 where (rec.user_id, rec.TRANSACTION_ID_CHILD) = (table1.USER_ID, table1.TRANSACTION_ID_PARENT)
)
select * from rec
order by USER_ID, TRANSACTION_ID_ROOT, TRANSACTION_ID_PARENT, TRANSACTION_ID_CHILD

USER_ID	TRANSACTION_ID_PARENT	TRANSACTION_ID_CHILD	TRANSACTION_ID_ROOT
1	0	1	1
1	1	2	1
1	2	3	1
1	0	11	11
1	11	12	11

小提琴