Question

我有一个递归表：

CREATE TABLE [dbo].[CATEGORIE](
            [id_categorie] [int] IDENTITY(1,1) NOT NULL,
            [Nom] [varchar](50) NOT NULL,
            [_Description] [varchar](2048) NULL,
            [id_categorie_1] [int] NULL,
            [id_codesActivite] [int] NULL,
    PRIMARY KEY CLUSTERED 
    (
        [id_categorie] ASC
    )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
    ) ON [PRIMARY]

ALTER TABLE [dbo].[CATEGORIE]  WITH CHECK ADD  CONSTRAINT 
[FK_CATEGORIE_id_categorie_1] FOREIGN KEY([id_categorie_1])
REFERENCES [dbo].[CATEGORIE] ([id_categorie])
GO

我想查询一个结果，该表的所有子项都在不同的列中，如下所示：

COL1 | COL2    | COL3
----------------------------
CAT1 |         |
CAT1 | SCAT1-1 |
CAT1 | SCAT1-1 | SSCAT1-1-1
CAT1 | SCAT1-1 | SSCAT1-1-2
CAT1 | SCAT1-1 | SSCAT1-1-3
CAT1 | SCAT1-2 |
CAT1 | SCAT1-2 | SSCAT1-2-1
CAT1 | SCAT1-2 | SSCAT1-2-2
CAT1 | SCAT1-2 | SSCAT1-2-3
CAT2 |         |
CAT2 | SCAT2-1 | 
CAT2 | SCAT2-2 | 
CAT3 |         |
CAT3 | SCAT3-1 | 
CAT3 | SCAT3-1 | SSCAT3-1-1
CAT3 | SCAT3-1 | SSCAT3-1-2

我认为我必须使用递归查询，例如此链接 https://sqlpro.developpez.com/cours/sqlserver/cte-recursives/

但是我找不到好的。

谢谢

编辑：

目前我对此代码有看法：

SELECT
  (SELECT Nom
   FROM dbo.CATEGORIE AS CAT3
   WHERE (id_categorie =
        (SELECT id_categorie_1
         FROM dbo.CATEGORIE AS CAT2
         WHERE (id_categorie = dbo.CATEGORIE.id_categorie_1)))) AS CAT1,

  (SELECT Nom
   FROM dbo.CATEGORIE AS CAT2
   WHERE (id_categorie = dbo.CATEGORIE.id_categorie_1)) AS CAT2,
   Nom AS CAT3,
   id_categorie
FROM dbo.CATEGORIE

可以，但是我需要在C＃代码中“移动”列

+------+--------+----------+--------------+
| CAT1 | CAT2   | CAT3     | id_categorie |
+------+--------+----------+--------------+
| NULL | NULL   | CAT1     | 392          |
+------+--------+----------+--------------+
| NULL | CAT1   | CAT1-1   | 393          |
+------+--------+----------+--------------+
| NULL | CAT1   | CAT1-2   | 394          |
+------+--------+----------+--------------+
| NULL | CAT1   | CAT1-3   | 395          |
+------+--------+----------+--------------+
| CAT1 | CAT1-1 | CAT1-1-1 | 396          |
+------+--------+----------+--------------+
| CAT1 | CAT1-1 | CAT1-1-2 | 397          |
+------+--------+----------+--------------+
| CAT1 | CAT1-2 | CAT1-2-1 | 398          |
+------+--------+----------+--------------+
| NULL | NULL   | CAT2     | 399          |
+------+--------+----------+--------------+
| NULL | NULL   | CAT3     | 400          |
+------+--------+----------+--------------+
| NULL | CAT3   | CAT3-1   | 401          |
+------+--------+----------+--------------+
| NULL | CAT3   | CAT3-2   | 402          |
+------+--------+----------+--------------+
| CAT3 | CAT3-2 | CAT3-2-1 | 403          |
+------+--------+----------+--------------+
| CAT3 | CAT3-2 | CAT3-2-2 | 404          |
+------+--------+----------+--------------+
| CAT3 | CAT3-2 | CAT3-2-3 | 405          |
+------+--------+----------+--------------+

Answer 1

尝试先构建父查询，然后再构建子查询，无论您要构建的深度如何。

例如，如果使用id_categorie_1 ='Parent'检索父对象，则父查询

Select id_categorie AS COL1 FROM CATEGORIE WHERE id_categorie_1='Parent'

对于父母（第一代）的孩子

Select catPARENT.id_categorie AS COL1, ISNULL(catChild1.id_categorie,'') AS COL2 
FROM CATEGORIE AS catPARENT LEFT JOIN CATEGORIE AS catChild1
ON catPARENT.id_categorie=catChild1.id_categorie_1
WHERE catPARENT.id_categorie_1='Parent'

问题的答案，孙子（第二代）

Select catPARENT.id_categorie AS COL1, ISNULL(catChild1.id_categorie,'') AS COL2
, ISNULL(catChild2.id_categorie,'') AS COL3
FROM CATEGORIE AS catPARENT LEFT JOIN CATEGORIE AS catChild1
ON catPARENT.id_categorie=catChild1.id_categorie_1 LEFT JOIN CATEGORIE AS catChild2
ON catChild1.id_categorie=catChild2.id_categorie_1 
WHERE catPARENT.id_categorie_1='Parent'

Answer 2

如果只需要三列，则可以执行以下操作：

select c.name, c1.nom, c2.nom
from categorie c left join
     categorie c1
     on c1.id_categorie_1 = c.id_categorie left join
     categorie c2
     on c2.id_categorie_1 = c1.id_categorie
where c.id_categorie_1 is null;

这将获得最高级别的类别和接下来的两个级别。对于此示例，无需进行递归查询。

Answer 3

使用Item i1 i2 i3 i4 User u1 NaN NaN NaN 2.0 u4 2.0 NaN NaN NaN u5 NaN 1.0 NaN 3.0。使用与每个子成员串联的recursive CTE。这最多支持递归9级。如果需要更多，请使用seq之类的数字代替01。还要在select语句中相应地添加列。

在不同列中有子项的递归查询

3 个答案: