sql-server - 如何在 SQL 中的单个列中显示递归 parentID

Question

这是表的示例结构:

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

我正在尝试编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，我将不胜感激:)

编辑::这是真正的查询，但它遵循相同的想法。它在线返回错误:marketparent.family + '^'+ t2.marketGroupName 因为它找不到 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 PostgreSQL

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)，在大多数现代 DBMS 上几乎不需要改动。

编辑 :

对于 SQL Server，您需要将标准的串联字符替换为 Microsoft 的非标准 + (并且您需要删除标准要求的 recursive 关键字，但由于某些奇怪的原因被 SQL Server 拒绝)

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;