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公司考核题-(ORACLE篇二)

时间:2015-02-18 17:40:09      阅读:329      评论:0      收藏:0      [点我收藏+]

标签:oracle

本篇主要考察的是算法--递归编程(树形的生成)--->本篇中的代码也可以使用ORACLE中的“CONNECT BY"来完成要求。

考核题目如下:

技术分享

程序如下:

1.创建表deptframwork

insert into deptframwork (DEPTID, UPPERID, DEPTNAME, DEPTNUM, CREATETIME, WHOLENAME, EMPLOYEE)
values (1, 0, ‘省公司电力本部‘, 1001, to_date(‘03-02-2010‘, ‘dd-mm-yyyy‘), ‘‘, ‘‘);

insert into deptframwork (DEPTID, UPPERID, DEPTNAME, DEPTNUM, CREATETIME, WHOLENAME, EMPLOYEE)
values (2, 1, ‘科技信息部‘, 1002, to_date(‘03-02-2010‘, ‘dd-mm-yyyy‘), ‘‘, ‘‘);

insert into deptframwork (DEPTID, UPPERID, DEPTNAME, DEPTNUM, CREATETIME, WHOLENAME, EMPLOYEE)
values (3, 1, ‘企业协会‘, 1003, to_date(‘03-02-2010‘, ‘dd-mm-yyyy‘), ‘‘, ‘‘);

insert into deptframwork (DEPTID, UPPERID, DEPTNAME, DEPTNUM, CREATETIME, WHOLENAME, EMPLOYEE)
values (4, 1, ‘经法部‘, 1004, to_date(‘03-02-2010‘, ‘dd-mm-yyyy‘), ‘‘, ‘‘);

insert into deptframwork (DEPTID, UPPERID, DEPTNAME, DEPTNUM, CREATETIME, WHOLENAME, EMPLOYEE)
values (5, 2, ‘法律事务处‘, 1005, to_date(‘03-02-2010‘, ‘dd-mm-yyyy‘), ‘‘, ‘‘);

insert into deptframwork (DEPTID, UPPERID, DEPTNAME, DEPTNUM, CREATETIME, WHOLENAME, EMPLOYEE)
values (6, 2, ‘体制改革处‘, 1006, to_date(‘03-02-2010‘, ‘dd-mm-yyyy‘), ‘‘, ‘‘);

2.创建表emptable
insert into emptable (EMID, EMNAME, PASSWD, DEPTNO)
values (1, ‘USER_ADMIN‘, ‘111111‘, 3);

insert into emptable (EMID, EMNAME, PASSWD, DEPTNO)
values (2, ‘USER_DEP‘, ‘111111‘, 3);

insert into emptable (EMID, EMNAME, PASSWD, DEPTNO)
values (3, ‘USER_SEA‘, ‘111111‘, 4);

insert into emptable (EMID, EMNAME, PASSWD, DEPTNO)
values (4, ‘USER_AAS‘, ‘111111‘, 4);

insert into emptable (EMID, EMNAME, PASSWD, DEPTNO)
values (5, ‘USER_TOC‘, ‘111111‘, 5);

insert into emptable (EMID, EMNAME, PASSWD, DEPTNO)
values (6, ‘USER_SOA‘, ‘111111‘, 5);

3.创建表
create table deptemployee(deptid,emname) as 
select et.deptno,et.emname from deptframwork df,deptemployee et 
where et.deptno=ef.deptid;

4存储过程
create or replace procedure dept_emp6 is
cursor db_cursor is select * from deptframwork order by deptid;
cursor db_cursor_e is select * from deptemployee;
whole_name varchar(100);
emps varchar(200):=null;
flag boolean:=true;
begin
  for r in db_cursor loop
    if r.upperid=0 then
      update deptframwork set wholename=r.deptname where deptid=r.deptid;
      dept_emp4(r.deptid,r.deptname);
    end if;
    for re in db_cursor_e loop
      if re.deptno=r.deptid then
         if flag then
           emps:=re.emname;
           flag:=false;
         else
           emps:=emps ||‘;‘ || re.emname;
         end if;
       end if;
    end loop;
    flag:=true;
    update deptframwork set employee=emps where deptid = r.deptid;
    emps:=null;
  end loop; 
end dept_emp6;

-- 递归生成树形结构-------------------------------------------------------------------------------
create or replace procedure dept_emp4(upper_id number,dept_name varchar) is
cursor db_cursor is select * from deptframwork where upperid=upper_id;
whole_name varchar(100);
begin
  for r in db_cursor loop
   whole_name:=dept_name || ‘>‘ || r.deptname;
   update deptframwork set wholename=whole_name where deptid=r.deptid;
   dept_emp4(r.deptid,whole_name);
  end loop;
end dept_emp4;

公司考核题-(ORACLE篇二)

标签:oracle

原文地址:http://blog.csdn.net/fengshuiyue/article/details/43877349

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