早些时候看到一篇，但是必须是在 id大于pid 的情况下才能使用。
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 ) srcORDER BY id 最后的结果如下：
ID 父ID 父到子之间级数 父到子路径------ ------ ------------ --------------- 1 0 0 ,0 2 1 1 ,0,1 3 1 1 ,0,1 4 2 2 ,0,1,2 5 2 2 ,0,1,2 6 3 2 ,0,1,3 7 6 3 ,0,1,3,6 8 0 0 ,0 9 8 1 ,0,8 10 8 1 ,0,8 11 8 1 ,0,8 12 9 2 ,0,8,9 13 9 2 ,0,8,9 14 12 3 ,0,8,9,12 15 12 3 ,0,8,9,12 16 15 4 ,0,8,9,12,15 17 15 4 ,0,8,9,12,15 18 3 2 ,0,1,3 19 2 2 ,0,1,2 20 6 3 ,0,1,3,6 21 8 1 ,0,8
1 SELECT *, 2 @Pn:=pid, 3 @path:=( 4 SELECT GROUP_CONCAT( 5 SUBSTRING_INDEX( 6 @Pn:= (SELECT CONCAT(pid,'|',nodename) FROM treenodes WHERE id=SUBSTRING_INDEX(@Pn,'|',1)), 7 '|', 8 -1 ) 9 ORDER BY id DESC SEPARATOR '-' )10 FROM treenodes WHERE @Pn IS NOT NULL ORDER BY id ASC ) 11 AS path,12 LENGTH (@path) - LENGTH (REPLACE(@path, '-', ''))+1 AS lv13 FROM treenodes;
结果如下
