SQL查询以获取层次结构中位于较高位置的所有记录的列表

表格：

+-----+------------+-------------+
| Id  | DocumentNo | ParentCCID  |
+-----+------------+-------------+
|  10 | CC001      | NULL        |
|  20 | CC002      | CC001       |
|  33 | CC003      | CC002       |
+-----+-------------+-------------+

CC003
CC002
CC001

select b2.documentno,b2.ParentCCID  from basicdetails  b1 
inner join basicdetails b2 on b1.documentno = b2.ParentCCID 
where b2.documentno='CC003' 

create table basicdetails2
  (
    id int identity,
    documentno varchar(30),
    parentccid varchar(30)
  )
    insert into basicdetails2 values('CC001', null)
    insert into basicdetails2 values('CC002', 'CC001')
    insert into basicdetails2 values('CC003', 'CC002')
    insert into basicdetails2 values('CC004', 'CC003')

传递给查询的值：

CC003

预期输出：

+-----+------------+-------------+
| Id  | DocumentNo | ParentCCID  |
+-----+------------+-------------+
|  10 | CC001      | NULL        |
|  20 | CC002      | CC001       |
|  33 | CC003      | CC002       |
+-----+-------------+-------------+

CC003
CC002
CC001

select b2.documentno,b2.ParentCCID  from basicdetails  b1 
inner join basicdetails b2 on b1.documentno = b2.ParentCCID 
where b2.documentno='CC003' 

create table basicdetails2
  (
    id int identity,
    documentno varchar(30),
    parentccid varchar(30)
  )
    insert into basicdetails2 values('CC001', null)
    insert into basicdetails2 values('CC002', 'CC001')
    insert into basicdetails2 values('CC003', 'CC002')
    insert into basicdetails2 values('CC004', 'CC003')

尝试失败：

+-----+------------+-------------+
| Id  | DocumentNo | ParentCCID  |
+-----+------------+-------------+
|  10 | CC001      | NULL        |
|  20 | CC002      | CC001       |
|  33 | CC003      | CC002       |
+-----+-------------+-------------+

CC003
CC002
CC001

select b2.documentno,b2.ParentCCID  from basicdetails  b1 
inner join basicdetails b2 on b1.documentno = b2.ParentCCID 
where b2.documentno='CC003' 

create table basicdetails2
  (
    id int identity,
    documentno varchar(30),
    parentccid varchar(30)
  )
    insert into basicdetails2 values('CC001', null)
    insert into basicdetails2 values('CC002', 'CC001')
    insert into basicdetails2 values('CC003', 'CC002')
    insert into basicdetails2 values('CC004', 'CC003')

注意：DocumentNo是唯一的主键。如果没有父记录，ParentCCID可能有空值

编辑：

+-----+------------+-------------+
| Id  | DocumentNo | ParentCCID  |
+-----+------------+-------------+
|  10 | CC001      | NULL        |
|  20 | CC002      | CC001       |
|  33 | CC003      | CC002       |
+-----+-------------+-------------+

CC003
CC002
CC001

select b2.documentno,b2.ParentCCID  from basicdetails  b1 
inner join basicdetails b2 on b1.documentno = b2.ParentCCID 
where b2.documentno='CC003' 

create table basicdetails2
  (
    id int identity,
    documentno varchar(30),
    parentccid varchar(30)
  )
    insert into basicdetails2 values('CC001', null)
    insert into basicdetails2 values('CC002', 'CC001')
    insert into basicdetails2 values('CC003', 'CC002')
    insert into basicdetails2 values('CC004', 'CC003')

---

您需要递归cte：

with cte as (
      select bd.documentno, bd.ParentCCID
      from basicdetails bd
      where bd.documentno = 'CC003'
      union all
      select cte.documentno, cte.ParentCCID
      from cte join
           basicdetails bd
           on bd.documentno = cte.ParentCCID
     )
select bd.documentno
from cte;

---

对戈登的答案（已经是+1）只是一个小小的改变

我喜欢跟踪水平，并看到每个记录的家长

示例

Declare @Fetch varchar(25) = 'CC003'

;with cte as (
      Select DocumentNo
            ,ParentCCDocumentNo
            ,Lvl=1
      From   YourTable 
      Where  DocumentNo=@Fetch
      Union  All
      Select R.DocumentNo
            ,R.ParentCCDocumentNo
            ,P.Lvl+1
      From   YourTable R
      Join   cte P on P.ParentCCDocumentNo = R.DocumentNo)
Select Lvl = Row_Number() over (Order By Lvl Desc)
      ,DocumentNo
      ,ParentCCDocumentNo
From cte
Order By 1 desc

返回

Lvl DocumentNo  ParentCCDocumentNo
3   CC003       CC002
2   CC002       CC001
1   CC001       NULL

---

为什么尝试失败了？所以成功了，但我想知道是否有一个非递归的解决方案？@SamuraiJack技术上是的，但他们可能会变得有点难看，特别是如果你有一个可变的深度层次结构。。您是否一次要查找多条路径？深度可能会有所不同，但一条路径只有一个直接父级record@SamuraiJack由于您的层次结构是可变的或参差不齐的，因此递归CTE是迄今为止最好的选择。这是另一个供您考虑的选项@SamuraiJack Methink认为，

join

应该是bd.documentno=cte.ParentCCID上的

。第二个
select
应该是
bd.documentno，bd.ParentCCID
，而不是
cte.
。