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
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