mssql server 树形结构的存储与查询实例
设计中常常会遇到需要存储树形结构,比如员工关系表、组织结构表,等等。
sql code
--测试数据 create table #employees( employeecode varchar(20) not null primary key clustered, reporttocode varchar(20) null) go insert into #employees values('a',null) insert into #employees values('b','a') insert into #employees values('c','a') insert into #employees values('d','a') insert into #employees values('e','b') insert into #employees values('f','b') insert into #employees values('g','c') insert into #employees values('h','d') insert into #employees values('i','d') insert into #employees values('j','d') insert into #employees values('k','j') insert into #employees values('l','j') insert into #employees values('m','j') insert into #employees values('n','k') go
/*
可能遇到的查询问题:
1. 员工'd'的所有直接下属
2. 员工'd'的所有2级以内的下属(包括直接下属和直接下属的下属)
3. 员工'n'的所有上级(按报告线顺序列出)
4. 员工@employeecode的所有@leveldown级以内的下属(@employeecode和@leveldown以变量传入)
declare @employeecode varchar(20), @leveldown int;
set @employeecode = 'd';
set @leveldown = 2;
5. 员工@employeecode的所有@levelup级以内的上级(@employeecode和@levelup以变量传入)
declare @employeecode varchar(20), @levelup int;
set @employeecode = 'n';
set @levelup = 2;
*/
--用递归cte实现员工树形关系表
with cte as(
select
employeecode,
reporttocode,
reporttodepth = 0,
reporttopath = cast('/' + employeecode + '/' as varchar(200))
from #employees
where reporttocode is null
union all
select
e.employeecode,
e.reporttocode,
reporttodepth = mgr.reporttodepth + 1,
reporttopath = cast(mgr.reporttopath + e.employeecode + '/' as varchar(200))
from #employees e
inner join cte mgr
on e.reporttocode = mgr.employeecode
)
select * from cte order by reporttopath
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