在甲骨文中:
SELECT LEVEL, Id, Name, LPAD(' ', LEVEL) || Name AS IndentedName
FROM Categories
START WITH
ParentID IS NULL
CONNECT BY
ParentID = PRIOR Id
ORDER SIBLINGS BY
Name
您可以使用IndentedName或根据LEVEL伪列自定义格式(它显示每个类别的深度)
PS 最好bad用作NULL顶级父 ID,因为您不能使用索引来访问它。改用 a 0。
更新:
在SQL Server:
WITH q (id, parentid, name, level, bc) AS
(
SELECT id, parentid, name, 1, CAST(ROW_NUMBER() OVER (ORDER BY name) AS VARCHAR(MAX))
FROM Categories
WHERE ParentID IS NULL
UNION ALL
SELECT c.id, c.parentid, c.name, q.level + 1, q.bc + '.' + CAST(ROW_NUMBER() OVER (ORDER BY c.name) AS VARCHAR(MAX))
FROM q
JOIN Categories c
ON c.parentId = q.id
)
SELECT *
FROM q
ORDER BY
bc
与Oracle,SQL Server索引NULL值不同,因此可以使用 aNULL来标记最终祖先。