关于SQL递归查询在不同数据库中的实现方法

比如表结构数据如下:

Table:Tree

ID Name ParentId

1 一级  0

2  二级  1

3  三级  2

4 四级  3

 

SQL SERVER 2005查询方法:

//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,还可用如下方法:

增加一列TID,类型设为:hierarchyid(这个是CLR类型,表示层级),且取消ParentId字段,变成如下:(表名为:Tree2)

TId    Id    Name

0x      1     一级
0x58     2    二级
0x5B40   3   三级
0x5B5E   4   四级

查询方法:

SELECT *,TId.GetLevel() as [level] FROM Tree2 --获取所有层级

DECLARE @ParentTree hierarchyid
SELECT @ParentTree=TId FROM Tree2 WHERE Id=2
SELECT *,TId.GetLevel()AS  [level] FROM Tree2 WHERE TId.IsDescendantOf(@ParentTree)=1 --获取指定的节点所有下级

DECLARE @ChildTree hierarchyid
SELECT @ChildTree=TId FROM Tree2 WHERE Id=3
SELECT *,TId.GetLevel()AS  [level] FROM Tree2 WHERE @ChildTree.IsDescendantOf(TId)=1 --获取指定的节点所有上级

可参见相关文章:http://blog.csdn.net/szstephenzhou/article/details/8277667

ORACLE中的查询方法:

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

可参见相关文章:http://blog.csdn.net/super_marioli/article/details/6253639

MYSQL 中的查询方法:

//定义一个依据ID查询所有父ID为这个指定的ID的字符串列表,以逗号分隔
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


//查询方法:
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

补充说明:上面这个方法在下查是没有问题,但在上查时会出现问题(详见博问:http://q.cnblogs.com/q/76375/),原因在于我的逻辑写错了,存在死循环,现已修正,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

这样递归查询就很方便了。

可参见相关文章:http://blog.csdn.net/jackiehome/article/details/6803978

说明:以上知识点均来源于网上,我这里只是做一个合计总结。

posted @ 2015-10-12 17:59  梦在旅途  阅读(3069)  评论(0编辑  收藏