SQL 嵌套查询

Question

我有一张如下表

id  name     dependency
-----------------------
1   xxxx     0
2   yyyy     1
3   zzzz     2 
4   aaaaaa   0
5   bbbbbb   4
6   cccccc   5

名单还在继续。我想从这个表中选择一组行，方法是在 SQL 的 where 子句中给出 0 依赖项的名称，直到它达到没有更多依赖项的条件。 (例如，第 1,2,3 行形成一个组，第 4,5,6 行是另一个组)。请帮助

Answer 1

由于您没有指定产品，我将使用 SQL 规范中可用的功能。在这种情况下，我使用了许多数据库产品支持的公用表表达式，包括 SQL Server 2005+ 和 Oracle(但不包括 MySQL):

With MyDependents As
    (
    Select id, name, 0 As level
    From MyTable
    Where dependency = 0
        And name = 'some value'
    Union All
    Select T.id, T.name, T.Level + 1
    From MyDependents As D
        Join MyTable As T
            On T.id = D.dependency
    )
Select id, name, level
From MyDependents

另一个不依赖于公用表表达式但确实假设最大深度级别(在这种情况下低于 0 级的两个级别)的解决方案将类似于

Select T1.id, T1.name, 0 As level
From MyTable As T1
Where T1.name = 'some value'
Union All
Select T2.id, T2.name, 1 
From MyTable As T1
    Join MyTable As T2
        On T2.Id = T1.Dependency
Where T1.name = 'some value'
Union All
Select T3.id, T3.name, 2
From MyTable As T1
    Join MyTable As T2
        On T2.Id = T1.Dependency
    Join MyTable As T3
        On T3.Id = T2.Dependency
Where T1.name = 'some value'