有關(guān)數(shù)據(jù)庫SQL遞歸查詢?cè)诓煌瑪?shù)據(jù)庫中的實(shí)現(xiàn)方法
本文給大家介紹有關(guān)數(shù)據(jù)庫SQL遞歸查詢?cè)诓煌瑪?shù)據(jù)庫中的實(shí)現(xiàn)方法,具體內(nèi)容請(qǐng)看下文。
比如表結(jié)構(gòu)數(shù)據(jù)如下:
Table:Tree
ID Name ParentId
1 一級(jí) 0
2 二級(jí) 1
3 三級(jí) 2
4 四級(jí) 3
SQL SERVER 2005查詢方法:
//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,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 = '$';
IF direction=1 THEN
SET sTempChd =cast(rootId as CHAR);
ELSEIF direction=2 THEN
SELECT cast(ParentId as CHAR) into sTempChd FROM Tree WHERE Id=rootId;
END IF;
WHILE sTempChd is not null DO
SET sTemp = concat(sTemp,',',sTempChd);
SELECT group_concat(id) INTO sTempChd FROM Tree where (direction=1 and FIND_IN_SET(ParentId,sTempChd)>0)
or (direction=2 and FIND_IN_SET(Id,sTempChd)>0);
END WHILE;
RETURN sTemp;
END
//查詢方法:
select * from tree where find_in_set(id,getChildLst(1,1));--下查
select * from tree where find_in_set(id,getChildLst(1,2));--上查
補(bǔ)充說明:上面這個(gè)方法在下查是沒有問題,但在上查時(shí)會(huì)出現(xiàn)問題,原因在于我的邏輯寫錯(cuò)了,存在死循環(huán),現(xiàn)已修正,新的方法如下:
CREATE DEFINER=`root`@`localhost` FUNCTION `getChildLst`(rootId int,direction int) RETURNS varchar(1000) CHARSET utf8
BEGIN
DECLARE sTemp VARCHAR(5000);
DECLARE sTempChd VARCHAR(1000);
SET sTemp = '$';
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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