Sql server Sql Server递归父/子表使用相同字段

我需要从同一张桌子上得到父母和所有孩子

运行此查询时：

select location, children  from lochierarchy where parent ='HSAGCF'

这一结果：

Location    Children
--------    ------------
HSAGCFSMF   1
HSAGEE      1
HSAGGAI     0
HSAGPO      1
HSAGSA      1
HSAGSACC    1
HSAGSAFR    0
HSAGSARV    0
HSAGSASG    0
HSAGSC      1
HSAGSD      1
HSAGSI      1
HSAGSO      1
HSAGSR      0
HSAGST      0
HSAGSTTO    0

当位置有子对象=1时，层次结构中有更多子对象

如何执行递归操作，在children=1的位置上获取该查询的每个位置，然后再次按如下方式运行该查询：

select location 
  from lochierarchy 
 where parent in ( 'HSAGCFSMF', 'HSAGEE', 'HSAGGAI', 'HSAGPO', 'HSAGSA',
                   'HSAGSACC','HSAGSAFR', 'HSAGSARV', 'HSAGSASG', 'HSAGSC',
                   'HSAGSD', 'HSAGSI', 'HSAGSO', 'HSAGSR', 'HSAGST', 
                   'HSAGSTTO', 'HSAGSV', 'HSAGU1', 'HSAGU2', 'HSAGU3', 'HSAGU4')


locations     Children
----------    ------------
HSAGCFSMF       1
HSAGEE          1
HSAGGAI         0
HSAGPO          1
HSAGSA          1
HSAGSACC        1
HSAGSAFR        0
HSAGSARV        0
HSAGSASG        0
HSAGSC          1
HSAGSD          1
HSAGSI          1
HSAGSO          1
HSAGSR          0
HSAGST          0
HSAGSTTO        0
HSAGSV          0

declare @parent varchar(20) = 'HSAGCF';

with cte as (
   select location, parent 
   from lochierarchy
   where location = @parent

   union all

   select c.location, c.parent
   from lochierarchy as c
   join cte as p
      on c.parent = p.location
)
select *
from cte;

---

需要重复此递归，直到所有位置的子位置都为0。如果使用SQL 2005或更高版本，则可以使用递归公共表表达式。大概是这样的：

select location 
  from lochierarchy 
 where parent in ( 'HSAGCFSMF', 'HSAGEE', 'HSAGGAI', 'HSAGPO', 'HSAGSA',
                   'HSAGSACC','HSAGSAFR', 'HSAGSARV', 'HSAGSASG', 'HSAGSC',
                   'HSAGSD', 'HSAGSI', 'HSAGSO', 'HSAGSR', 'HSAGST', 
                   'HSAGSTTO', 'HSAGSV', 'HSAGU1', 'HSAGU2', 'HSAGU3', 'HSAGU4')


locations     Children
----------    ------------
HSAGCFSMF       1
HSAGEE          1
HSAGGAI         0
HSAGPO          1
HSAGSA          1
HSAGSACC        1
HSAGSAFR        0
HSAGSARV        0
HSAGSASG        0
HSAGSC          1
HSAGSD          1
HSAGSI          1
HSAGSO          1
HSAGSR          0
HSAGST          0
HSAGSTTO        0
HSAGSV          0

declare @parent varchar(20) = 'HSAGCF';

with cte as (
   select location, parent 
   from lochierarchy
   where location = @parent

   union all

   select c.location, c.parent
   from lochierarchy as c
   join cte as p
      on c.parent = p.location
)
select *
from cte;

---

如果您想要树中叶节点的路径，我将此作为练习留给读者。

现在您的问题还不够清楚。它没有解释父项和子项之间的链接是什么（除了0，1字段）。是位置字段的字母吗？或者别的什么？请更好地解释一下。@JorgeCampos:是的，只是没有明确说明。查看OP最初运行的查询：

select location，children from lochierarchy where parent='HSAGCF'

。这意味着有一个父列包含当前行的父列。哦，你是对的@BenThul，谢谢。