SQL如何实现MYSQL的递归查询(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

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