路径枚举模型,如何获得上级?

时间:2011-05-26 13:55:04

标签: sql sql-server-2008

考虑下面的表格数据:

emp_name VARCHAR(25), path VARCHAR(150)
Albert                /Albert
John                  /Albert/John
Chuck                 /Albert/Chuck
Tom                   /Albert/John/Tom
Frank                 /Frank

我想得到汤姆的上级名单:

John
Albert

(可以包括汤姆)

是否可以不分割路径,然后使用序列表(只有我找到的方式)?

DB是Sql server 2008 R2

1 个答案:

答案 0 :(得分:1)

您可以使用递归CTE分割层次结构字符串值。

declare @T table 
(
  ID int identity,
  emp_name varchar(25),
  [path] varchar(150)
)

insert into @T values
('Albert', 'Albert'),
('John',   'Albert/John'),
('Chuck',  'Albert/Chuck'),
('Tom',    'Albert/John/Tom'),
('Frank',  'Frank')

declare @EmpName varchar(25) = 'Tom'

;with cte(Sort, P1, P2, [path]) as
(
  select 1,
         1,
         charindex('/', [path]+'/', 1),
         [path]
  from @T
  where emp_name = @EmpName  
  union all
  select Sort+1,
         P2+1,
         charindex('/', [path]+'/', C.P2+1),
         [path]
  from cte as C
  where charindex('/', [path]+'/', C.P2+1) > 0
)
select substring([path], P1, P2-P1)
from cte
order by Sort

结果:

(No column name)
Albert
John
Tom

在此处测试查询:http://data.stackexchange.com/stackoverflow/q/101383/

你可以尝试的另一件事

select T2.emp_name
from @T as T1
  inner join @T as T2
    on '/'+T1.[path]+'/' like '%/'+T2.emp_name+'/%' and
       T2.emp_name <> @EmpName  
where T1.emp_name = @EmpName

http://data.stackexchange.com/stackoverflow/q/101518/get-hierarchy-with-join-using-like