[转]Mysql中模拟递归查询

众所周知,目前的mysql版本中并不支持直接的递归查询,但是通过递归到迭代转化的思路,还是可以在一句SQL内实现树的递归查询的。这个得益于Mysql允许在SQL语句内使用@变量。以下是示例代码。

创建表格

CREATE TABLE `treenodes` (
`id` int , -- 节点ID
`nodename` varchar (60), -- 节点名称
`pid` int  -- 节点父ID
);

插入测试数据

INSERT INTO `treenodes` (`id`, `nodename`, `pid`) VALUES
('1','A','0'),('2','B','1'),('3','C','1'),
('4','D','2'),('5','E','2'),('6','F','3'),
('7','G','6'),('8','H','0'),('9','I','8'),
('10','J','8'),('11','K','8'),('12','L','9'),
('13','M','9'),('14','N','12'),('15','O','12'),
('16','P','15'),('17','Q','15'),('18','R','3'),
('19','S','2'),('20','T','6'),('21','U','8');

查询语句

SELECT id AS ID,pid AS 父ID ,levels AS 父到子之间级数, paths AS 父到子路径 FROM (
    SELECT id, pid,
        @le:= IF (pid = 0 ,0,
        IF( LOCATE( CONCAT('|',pid,':'),@pathlevel) > 0  ,
					SUBSTRING_INDEX( SUBSTRING_INDEX(@pathlevel,CONCAT('|',pid,':'),-1),'|',1) +1
					,@le+1) ) levels,
        @pathlevel:= CONCAT(@pathlevel,'|',id,':', @le ,'|') pathlevel,
        @pathnodes:= IF( pid =0,',0',
        CONCAT_WS(',',
        IF( LOCATE( CONCAT('|',pid,':'),@pathall) > 0  ,
					SUBSTRING_INDEX( SUBSTRING_INDEX(@pathall,CONCAT('|',pid,':'),-1),'|',1)
					,@pathnodes ) ,pid  ) )paths,
        @pathall:=CONCAT(@pathall,'|',id,':', @pathnodes ,'|') pathall
    FROM  treenodes,
    (SELECT @le:=0,@pathlevel:='', @pathall:='',@pathnodes:='') vv
    ORDER BY  pid,id
) src
ORDER BY id

Comments are closed.