SQL递归查询实现跟帖盖楼效果

网易新闻的盖楼乐趣多,某一天也想实现诸如网易新闻跟帖盖楼的功能,无奈技术不佳(基础不牢),网上搜索了资料才发现SQL查询方法有一种叫递归查询,整理如下:

一、查询出 id = 1 的所有子结点

with my1 as
(select * from table where id = 1  union all 
select table.* from my1, table 
where my1.id = table.fatherId) 
select * from my1    

结果包含1这条记录,如果不想包含,可以在最后加上:where id <> 1

二、查询出 id = 2 的所有父结点

with my1 as
(select * from table where id = 2  union all select table.* 
from my1, table where my1.fatherId = table.id ) 
select * from my1;  

三、删除 id = 1 的所有子结点(包括id = 1结点)

with my1 as
(select * from table where id = 1  union all select table.* 
from my1, table where my1.id = table.fatherId ) 
delete from table where exists (select id from my1 
where my1.id = table.id) 

SQL递归查询实现跟帖盖楼效果

本文摘自木庄网络博客>>SQL递归查询实现跟帖盖楼效果

posted @ 2016-11-19 13:09  木庄  阅读(821)  评论(0编辑  收藏  举报