//上查
with tmpTree
as
(
  select * from Tree where Id=2
  union all
  select p.* from tmpTree inner join Tree p on p.Id=tmpTree.ParentId
)
select * from tmpTree
 
//下查
with tmpTree
as
(
  select * from Tree where Id=2
  union all
  select s.* from tmpTree inner join Tree s on s.ParentId=tmpTree.Id
)
select * from tmpTree

SQL SERVER 2008及以后版本，還可用如下方法：

增加一列TID,類型設(shè)為：hierarchyid(這個(gè)是CLR類型，表示層級(jí))，且取消ParentId字段，變成如下：（表名為：Tree2）

TId　　　　Id　　　　Name

0x 　　　　1 　　　　一級(jí)
0x58 　　2　　　二級(jí)
0x5B40 　　3 　　三級(jí)
0x5B5E 　　4 　　四級(jí)

查詢方法：

SELECT *,TId.GetLevel() as [level] FROM Tree2 --獲取所有層級(jí)
DECLARE @ParentTree hierarchyid
SELECT @ParentTree=TId FROM Tree2 WHERE Id=2
SELECT *,TId.GetLevel()AS [level] FROM Tree2 WHERE TId.IsDescendantOf(@ParentTree)=1 --獲取指定的節(jié)點(diǎn)所有下級(jí)
DECLARE @ChildTree hierarchyid
SELECT @ChildTree=TId FROM Tree2 WHERE Id=3
SELECT *,TId.GetLevel()AS [level] FROM Tree2 WHERE @ChildTree.IsDescendantOf(TId)=1 --獲取指定的節(jié)點(diǎn)所有上級(jí)

ORACLE中的查詢方法：

SELECT *
FROM Tree
START WITH Id=2
CONNECT BY PRIOR ID=ParentId --下查
SELECT *
FROM Tree
START WITH Id=2
CONNECT BY ID= PRIOR ParentId --上查

MYSQL 中的查詢方法：

//定義一個(gè)依據(jù)ID查詢所有父ID為這個(gè)指定的ID的字符串列表，以逗號(hào)分隔
CREATE DEFINER=`root`@`localhost` FUNCTION `getChildLst`(rootId int,direction int) RETURNS varchar(1000) CHARSET utf8
BEGIN
 DECLARE sTemp VARCHAR(5000);
  DECLARE sTempChd VARCHAR(1000);
  SET sTemp = '$';
  IF direction=1 THEN
   SET sTempChd =cast(rootId as CHAR);
  ELSEIF direction=2 THEN
   SELECT cast(ParentId as CHAR) into sTempChd FROM Tree WHERE Id=rootId;
  END IF;
  WHILE sTempChd is not null DO
    SET sTemp = concat(sTemp,',',sTempChd);
    SELECT group_concat(id) INTO sTempChd FROM Tree where (direction=1 and FIND_IN_SET(ParentId,sTempChd)>0)
    or (direction=2 and FIND_IN_SET(Id,sTempChd)>0);
  END WHILE;
RETURN sTemp;
END
//查詢方法：
select * from tree where find_in_set(id,getChildLst(1,1));--下查
select * from tree where find_in_set(id,getChildLst(1,2));--上查

補(bǔ)充說(shuō)明：上面這個(gè)方法在下查是沒(méi)有問(wèn)題，但在上查時(shí)會(huì)出現(xiàn)問(wèn)題，原因在于我的邏輯寫錯(cuò)了，存在死循環(huán)，現(xiàn)已修正，新的方法如下：

CREATE DEFINER=`root`@`localhost` FUNCTION `getChildLst`(rootId int,direction int) RETURNS varchar(1000) CHARSET utf8
BEGIN
 DECLARE sTemp VARCHAR(5000);
  DECLARE sTempChd VARCHAR(1000);
  SET sTemp = '$';
  SET sTempChd =cast(rootId as CHAR);
  
  IF direction=1 THEN
  WHILE sTempChd is not null DO
    SET sTemp = concat(sTemp,',',sTempChd);
    SELECT group_concat(id) INTO sTempChd FROM Tree where FIND_IN_SET(ParentId,sTempChd)>0;
  END WHILE;
  ELSEIF direction=2 THEN
  WHILE sTempChd is not null DO
    SET sTemp = concat(sTemp,',',sTempChd);
    SELECT group_concat(ParentId) INTO sTempChd FROM Tree where FIND_IN_SET(Id,sTempChd)>0;
  END WHILE;
  END IF;
RETURN sTemp;
END

這樣遞歸查詢就很方便了。

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