一句SQL实现MYSQL的递归查询
2015-07-31 10:48 dukou 阅读(44189) 评论(3) 编辑 收藏 举报众所周知,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