如下一张表test:id name pid----------- ---------- -----------1 电器 NULL2 家电 13 冰箱 24 洗衣机 25 电脑 16 笔记本 57 平板 58 组装机 79 品牌机 7--查询电脑的所有子节点. 可采用标准sql的with实现递归查询: with subRecord(id,name,pid) as ( select id,name,pid from test where id = 5 union all select test.id
用标准sql的with实现递归查询(sql2005以上肯定支持,sql2000不清楚是否支持): with subqry(id,name,pid) as ( select id,name,pid from test1 where id = 5 --指定id union all select test1.id,test1.name,test1.pid from test1,subqry where test1.pid = subqry.id)select* from subqry;
Oracle树形结构递归查询 在Oracle中,对于树形查询可以使用start with ... connect by select * from treeTable start with id='1' connect by id = prior parent_id; 若将一个树状结构存储在一张表里,需要在表中存入两个字段ID和PARENTID,表示每一条记录的parent是谁. table: treeTable 1.从根节点遍历子节点.:(一整棵树) select * from treeTab
1.所示案例数据表结构设计如下所示: 2.案例数据如下所示: 3.mysql查询语句可以查询出父级目录信息: 注意:自己的数据表表名称,切记手动修改,字段名称(特别注意id,parent_id字段名称,不然肯定查询不出来的). SELECT T2.* FROM ( SELECT @r AS _id, (SELECT @r := parent_id FROM catelog WHERE id = _id) AS parent_id, @l := @l + AS lvl FROM (SELECT @
SELECT t3.college_code FROM ( SELECT t1.college_code, IF ( find_in_set( t1.parent_org_code, , ) AS ischild FROM ( ORDER BY parent_org_code, college_code ) t1, ( ' college_code ) t2 ) t3 WHERE t3.ischild != 0 表结构: t_college: id, college_code (机构编码), p
stackoverflow的解决方案,亲测有效: SELECT * FROM person WHERE department IN (SELECT department_id FROM department WHERE department_id = UNION (SELECT department_id FROM (SELECT * FROM department ORDER BY parent_id,department_id) depart_sorted, (SELECT @pv := )
定义一个函数 ) CHARSET utf8 BEGIN ); ); SET sTemp = '$'; SET sTempChd =cast(rootId as CHAR); WHILE sTempChd is not null DO SET sTemp = concat(sTemp,',',sTempChd); ORDER BY sort, column_id; END WHILE; RETURN sTemp; END 使用: select t1.article_id, t1.sort,t1.s
首先进入到mysql里 show databases; 选择数据库 use xxxcms; 查询数据库下的表结构 show create table 表名; 这样看着不太好可以后面加\G show create table 表名\G; 如上所示并没有索引创建 下面来查询一下索引 show indexes from 表名\G; 图11 以上返回值的官方介绍 http://dev.mysql.com/doc/refman/5.6/en/show-index.html 主要Key_name 索引的名字