一句 SQL 实现 MYSQL 的递归查询

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
​