递归 cte 来拯救......
创建并填充示例表(请在您以后的问题中保存我们这一步):
DECLARE @T as table
(
id int,
name varchar(100),
parent_id int
)
INSERT INTO @T VALUES
(1, 'A', NULL),
(2, 'A.1', 1),
(3, 'A.2', 1),
(4, 'A.1.1', 2),
(5, 'B', NULL),
(6, 'B.1', 5),
(7, 'B.1.1', 6),
(8, 'B.2', 5),
(9, 'A.1.1.1', 4),
(10, 'A.1.1.2', 4)
CTE:
;WITH CTE AS
(
SELECT id, name, name as path, parent_id
FROM @T
WHERE parent_id IS NULL
UNION ALL
SELECT t.id, t.name, cast(cte.path +','+ t.name as varchar(100)), t.parent_id
FROM @T t
INNER JOIN CTE ON t.parent_id = CTE.id
)
查询:
SELECT id, name, path
FROM CTE
结果:
id name path
1 A A
5 B B
6 B.1 B,B.1
8 B.2 B,B.2
7 B.1.1 B,B.1,B.1.1
2 A.1 A,A.1
3 A.2 A,A.2
4 A.1.1 A,A.1,A.1.1
9 A.1.1.1 A,A.1,A.1.1,A.1.1.1
10 A.1.1.2 A,A.1,A.1.1,A.1.1.2
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