如何在SQL中的单个列中显示递归parentID

Question

以下是表的示例结构:

ID    Name     ParentID
-----------------------
1     Ancestor      NULL
2     GrandFather   1
3     GrandMother   1
4     Child         3

我正在尝试写一个会返回的查询

ID     Name        Family
----------------------------
 1     Ancestor 
 2     GrandFather Ancestor
 3     GrandMother Ancestor
 4     Child       Ancestor^GrandMother

棘手的部分是我想以自上而下的顺序显示所有行的族.

如果有人能指出我正确的方向,将不胜感激:)

编辑::这是真正的查询,但它遵循相同的想法.它在行上返回错误:marketparent.family +'^'+ t2.marketGroupName因为它无法找到marketparent

WITH marketparent ( marketGroupID,parentGroupID, marketGroupName,family)
AS
(
SELECT marketGroupID,
       parentGroupID,
       marketGroupName,
       '' as family 
 FROM EVE.dbo.invMarketGroups
 WHERE parentGroupID IS NULL
UNION ALL

    SELECT t2.parentGroupID,
     t2.marketGroupID,
     t2.marketGroupName,
     marketparent.family + '^'+ t2.marketGroupName
     FROM EVE.dbo.invMarketGroups as t2
     INNER JOIN marketparent as mp
     ON mp.marketGroupID = t2.parentGroupID
)

-- Statement using the CTE

SELECT TOP 10 *
FROM marketparent;

Answer 1

你没有指定你的DBMS,所以我假设P​​ostgreSQL

WITH RECURSIVE fam_tree (id, name, parent, family) as 
(
  SELECT id, 
         name, 
         parentid, 
         ''::text as family
  FROM the_unknown_table
  WHERE parent IS NULL

  UNION ALL

  SELECT t2.id, 
         t2.name, 
         t2.parentid, 
         fam_tree.family || '^' || t2.name
  FROM the_unknown_table t2 
     INNER JOIN fam_tree ON fam_tree.id = t2.parentid
)
SELECT *
FROM fam_tree;

这是标准的SQL(除了::text类型转换),它应该在大多数现代DBMS上进行很少的更改.

编辑:

对于SQL Server,您需要使用Microsoft的非标准替换标准连接字符+(并且您需要删除recursive标准所需的关键字,但出于某些由SQL Server拒绝的奇怪原因)

WITH fam_tree (id, name, parent, family) as 
(
  SELECT id, 
         name, 
         parentid, 
         '' as family
  FROM the_unknown_table
  WHERE parent IS NULL

  UNION ALL

  SELECT t2.id, 
         t2.name, 
         t2.parentid, 
         fam_tree.family + '^' + t2.name
  FROM the_unknown_table t2 
     INNER JOIN fam_tree ON fam_tree.id = t2.parentid
)
SELECT *
FROM fam_tree;