时间:2021-07-01 10:21:17 帮助过:4人阅读
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
一句SQL实现MYSQL的递归查询
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