WITH B (FATHER,SON,ID,ALLINFO) AS (SELECT RTRIM(LTRIM(CHAR(A.ID)))|| CHAR(ROW_NUMBER() OVER(PARTITION BY ID)) AS FATHER, ) AS SON, A.ID, )) FROM T_DETAIL A UNION ALL SELECT C.FATHER,C.SON,C.ID, RTRIM(LTRIM(B.ALLINFO))||','||RTRIM(LTRIM(CHAR(C.INFO)))
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,
SQL递归查询(with cte as) with cte as( select Id,Pid,DeptName,0 as lvl from Department where Id = 2 union all select d.Id,d.Pid,d.DeptName,lvl+1 from cte c inner join Department d on c.Id = d.Pid)select * from cte 1 表结构 Id Pid
比如表结构数据如下: Table:Tree ID Name ParentId 1 一级 0 2 二级 1 3 三级 2 4 四级 3 SQL SERVER 2005查询方法: //上查 with tmpTree as ( select * from Tree where Id=2 union all select p.* from tmpTree inner join Tree p on p.Id=tmpTree.ParentId ) select * from tmpTree //下查
今天用到了sql的递归查询.递归查询是CTE语句with xx as(....)实现的. 假如表Category数据如下. 我们想查找机枪这个子分类极其层次关系(通过子节点,查询所有层级节点).以下是查询语句 WITH tt AS ( SELECT CategoryId,Name,Parent,0 level FROM dbo.Category WHERE CategoryId=15 --定位点成员 UNION ALL SELECT c.CategoryId,c.Name,c.Parent,tt
1.easyUI 的tree 修改节点: 我需要:切换语言状态,英文下, 修改根节点文本,显示英文. 操作位置:在tree的显示 $('#tree').tree(),onLoadSuccess事件方法中,参数:node,data(详见easyUI文档) 代码: var t = $(this); if (data) { var troot = t
with cte as ( select Id,Pid,DeptName,0 as lvl from Department where Id = 2 union all select d.Id,d.Pid,d.DeptName,lvl+1 from cte c inner join Department d on c.Id = d.Pid ) select * from cte 表结构 Id Pid DeptName ----------- ----------- ---------------
--> 生成测试数据表: [tb] IF OBJECT_ID('[Users]') IS NOT NULL DROP TABLE [Users] GO CREATE TABLE [Users] ([userid] [int],[username] [nvarchar](10),[parentUserId] [int],[parentUserName] [nvarchar](10)) INSERT INTO [Users] SELECT '1','admin','0',NULL UNION
--------------------01.向上查找所有父节点-----------------WITH TEMP AS (SELECT * FROM CO_Department WHERE ID=11 --表的IDUNION ALL SELECT T0.* FROM TEMP,CO_Department T0 WHERE TEMP.ParentID=T0.ID --父级ID==子级ID)SELECT * FROM TEMP; 如图:根据"测试组"查找所有父节点 执行结果
Declare @Id Int Set @Id = 0; ---在此修改父节点 With RootNodeCTE(D_ID,D_FatherID,D_Name,lv) As ( Select D_ID,D_FatherID,D_Name,0 as lv From [LFBMP.LDS].[dbo].[LDS.Dictionary] Where D_FatherID In (@Id) Union All Select [LFBMP.LDS].[dbo].[LDS.Dictionary].D_ID,
数据库设计中经常碰到父子节点的关系结构,经常需要找到某个节点的根,或者某个节点的所有子节点,一般做法都是在业务层做递归的方式实现,或者数据库存储过程实现.但其实SQLServer提供的CTE可以很好的简化我们的工作,非常方便的实现这一功能. 例子: 1.正向递归,找某个节点下的所有子节点 with t as--如果CTE前面有语句,需要用分号隔断(selectId,ParentId,Name from WMS_Org whereId'union all select r1.Id,r1.Paren
WITH T(emp_no, name, dept_no, the_level, path,path1,manager_id) AS( SELECT emp_no, name, dept_no ,1 AS the_level ,'\'||name path ,'\'||manager_id path1 ,manager_id FROM BASE_HR_EMP where emp_no='10002241' UNION ALL SELECT e.emp_no, e.name, e.dept_no