sql - 获取项目的所有祖先的 SQL 递归查询

Question

ID       parent_id   name
---------------------
1        2            first 
2        4            second
3        3            third
4        5            fourth
5        -           fifth

祖先名单 first应该是 (2, 4, 5)

Answer 1

with name_tree as (
   select id, parent_id, name
   from the_unknown_table
   where id = 1 -- this is the starting point you want in your recursion
   union all
   select c.id, c.parent_id, c.name
   from the_unknown_table c
     join name_tree p on p.parent_id = c.id  -- this is the recursion
) 
select *
from name_tree
where id <> 1; -- exclude the starting point from the overall result

SQLFiddle：http ://sqlfiddle.com/#!3/87d0c/1

Answer 2

你可以使用这样的东西：

with parents as 
(
  select ID, parent_ID
  from t
  where parent_ID is not null
  union all 
  select p.ID, t.parent_ID
  from parents p
    inner join t on p.parent_ID = t.ID
      and t.parent_ID is not null
      and t.ID <> t.parent_ID
)
select *
  , parents = '(' + stuff
    (
      (
        select ', ' + cast(p.parent_ID as varchar(100))
        from parents p 
        where t.ID = p.ID
        for xml path('')
      ), 1, 2, ''
    ) + ')'
from t
order by ID

SQL Fiddle 与演示。

这结合了两种非常常见的 T-SQL 技术——使用CTE获取层次结构和使用 FOR XML PATH 获取 CSV 列表。