Question

我有下表：

Account_Number  Parent_Account  Child_Account
R003247         R000355         R000002
R000355         NULL            R003247
R000002         R003247         NULL
R004853         NULL            R028636
R004853         NULL            R028638
R004853         NULL            R028637
R028636         R004853         NULL
R028638         R004853         NULL
R028637         R004853         NULL

可以加载：

create table dbo.temptable
(Account_Number varchar(10),
Parent_Account varchar(10),
Child_Account varchar(10))

insert into dbo.temptable
values
('R003247','R000355','R000002'),
('R000355',NULL,'R003247'),
('R000002','R003247',NULL),
('R004853',NULL,'R028636'),
('R004853',NULL,'R028638'),
('R004853',NULL,'R028637'),
('R028636','R004853',NULL),
('R028638','R004853',NULL),
('R028637','R004853',NULL)

此表表示帐号的拆分和重新分配。它与跟踪分裂和地块组合有关。

如上表所示，前三行应该组合在一起，因为它来自R000355 - ＆gt; R003247 - ＆gt; R000002

如上表所示，最后6个也应该进行分组，因为它显示R004853被分为三个R028636，R028637，R028638。< / p>

我尝试了很多这样的变体：

SELECT CE.*,TT.ID
FROM dbo.temptable CE 
INNER JOIN
    (
    SELECT ACCOUNT_NUMBER,ROW_NUMBER() OVER (ORDER BY ACCOUNT_NUMBER) AS ID
    FROM(
    SELECT DISTINCT ACCOUNT_NUMBER FROM dbo.temptable where Child_Account is not null)AA
    )TT 
ON TT.ACCOUNT_NUMBER = CE.Account_Number OR TT.Account_Number = CE.Child_Account

哪一个人：

Account_Number  Parent_Account  Child_Account   ID
R000355         NULL            R003247         1
R003247         R000355         R000002         2
R000355         NULL            R003247         2
R004853         NULL            R028636         3
R004853         NULL            R028638         3
R004853         NULL            R028637         3

当我真正需要的是：

Account_Number  Parent_Account  Child_Account   ID
R000355         NULL            R003247         1
R003247         R000355         R000002         1
R000002         R003247         NULL            1
R004853         NULL            R028636         2
R004853         NULL            R028638         2
R004853         NULL            R028637         2
R028636         R004853         NULL            2
R028638         R004853         NULL            2
R028637         R004853         NULL            2

Answer 1

这个问题没什么可怕的，一旦你跳进去给自己的手腕隧道在手机上打字过多。这只是一个略微修改的标准递归分层查询问题。值得注意的是，递归中的连接条件是当前帐号是某个父帐户的子帐号。至于编号，我们只使用DENSE_RANK而不是顶级父母。

WITH cte AS (
    SELECT m.*, DENSE_RANK() OVER (ORDER BY m.Account_Number) AS pos
    FROM temptable m
    WHERE Parent_Account IS NULL
    UNION ALL
    SELECT m.*, cte.pos
    FROM temptable m
    INNER JOIN cte
        ON m.Account_Number = cte.Child_Account
)

SELECT *
FROM cte
ORDER BY pos;

Demo

注意：我非常赞赏这里精彩的答案written by @Quassnoi。

创建链接ID

1 个答案:

Demo