oracle pl/sql 几道基础的编程题

今天学了pl/sql基础，做了几道基础的编程题，做个笔记，方便复习。如果有做错的地方或者有待改进的地方，欢迎指正，不胜感激。
练习参考的数据库表为oracle中自带的employees表。

1.求1~100之间素数

declare

  m_flag boolean := true;

begin

  for i in 1 .. 100 loop

    for j in 2 .. i - 1 loop

      if mod(i, j) = 0 then

        m_flag := false;

      end if;

    end loop;

    if m_flag then

      dbms_output.put_line(i);

    end if;

    m_flag := true;

  end loop;

end;

2.对所有JOB_ID为IT_PROG 薪水增加一倍.

declare

    cursor cur_raise_it is select * from employees e where e.job_id='IT_PROG' for update;

begin

    for i in cur_raise_it loop

        update employees e set e.salary=e.salary*2 where current of cur_raise_it;

    end loop;

end;

3.编写一个函数以检查所指定雇员的薪水是否有效范围内。不同职位的薪水范围为：

 JOB_ID  Range

 IT_PROG 5500-8500

 ST_CLERK 2501-3500

 SA_REP 7000-8500

 Others No operation.

 如果薪水在此范围内，则显示消息"Salary is OK"，否则，更新薪水为该范围内的最小值。

create or replace function check_sal(emp_no employees.employee_id%type)

  return varchar2 as

  v_job employees.job_id%type;

  v_sal employees.salary%type;

  v_msg varchar2(100) := 'Salary Is Ok';

begin

  select e.job_id, e.salary

    into v_job, v_sal

    from employees e

   where e.employee_id = emp_no;

  if v_job = 'IT_PROG' then

    if v_sal not between 5500 and 8500 then

      update employees e set e.salary = 8500;

      v_msg := 'salary changes from ' || v_sal || ' to' || ' 5500';

    end if;

  elsif v_job = 'ST_CLERK' then

    if v_sal not between 2501 and 3500 then

      update employees e set e.salary = 3500;

      v_msg := 'salary changes from ' || v_sal || ' to' || ' 2501';

    end if;

  elsif v_job = 'SA_REP' then

    if v_sal not between 7000 and 8500 then

      update employees e set e.salary = 8500;

      v_msg := 'salary changes from ' || v_sal || ' to' || ' 7000';

    end if;

  else

    null;

  end if;

  return v_msg;

end;

--调用示例

declare v_msg varchar2(100);

begin

v_msg:=check_sal(emp_no=>102);

dbms_output.put_line(v_msg);

end;

4.对名字以"A"或"H"开始的所有雇员按他们的基本薪水的10%加薪。

declare

cursor cur_sal_raise is select * from employees e w
4000
here substr(e.last_name,1,1) in('A','H');

begin

    for emp_record in cur_sal_raise loop

        update employees e set e.salary = e.salary * 1.1 WHERE CURRENT OF emp_record;

    end loop;

end;

5.编写一PL/SQL，以提升两个资格最老的"职员"为"高级职员"。（按工作时间算）

declare

  cursor cur_raise_old is

     select employee_id,

            first_name,

            last_name,

            email,

            job_id,

            hire_date,

            salary

       from (select e.employee_id,

                    e.job_id,

                    e.first_name,

                    e.last_name,

                    e.email,

                    e.hire_date,

                    e.salary,

                    months_between(sysdate, e.hire_date) months

               from employees e

              order by months desc) t

      where rownum <= 2 ;

begin

      for i in cur_raise_old loop

         update employees e set e.job_id ='高职' where i.employee_id =e.employee_id;

      end loop;

end;

6.编写存储过程检查如果已经雇佣该雇员超过60个月，则给他额外加薪3000.

create or replace procedure pro_raise_old(emp_no in number) is

       v_hiredate employees.hire_date%type;

       v_sal employees.salary%type;

begin

       select e.hire_date,e.salary into v_hiredate,v_sal from employees e where e.employee_id=emp_no;

       if months_between(sysdate,v_hiredate)>60 then v_sal:=v_sal+3000;

       end if;

       update employees e set e.salary=v_sal where e.employee_id=emp_no;

end;
--调用示例

begin

   pro_raise_old(emp_no=>103);

end;

7.编程实现对所有雇员按他们基本薪水的10%加薪，如果所增加的薪水大于1000，则取消加薪.

declare

  cursor cur_raise_two is

    select * from employees for update;

begin

  for i in cur_raise_two loop

    if i.salary * 0.1 > 1000 then

      null;

    else

      update employees e

         set e.salary = e.salary*1.1

       where current of cur_raise_two;

    end if;

  end loop;

end;

8.A单价49，B单价39，C单价17，现有2000元，要购买A,B,C，请给出花费最接近2000元（<=2000元）的购买计划，打印出来。（今天上课时的课堂练习）

DECLARE

  a            NUMBER := 49;

  b            NUMBER := 39;

  c            NUMBER := 17;

  total_amount NUMBER := 2000;

  a_qty        NUMBER;

  b_qty        NUMBER;

  c_qty        NUMBER;

  a_amount     number;

  b_amount     number;

  c_amount     number;

  l_max_amount NUMBER := 0;

BEGIN

  FOR i IN 0 .. trunc(total_amount / a) LOOP

    FOR j IN 0 .. trunc(total_amount / b) LOOP

      FOR k IN 0 .. trunc(total_amount / c) LOOP

        IF a * i + b * j + c * k >= l_max_amount AND

           a * i + b * j + c * k <= total_amount THEN

          l_max_amount := a * i + b * j + c * k;

          a_amount     := i;

          b_amount     := j;

          c_amount     := k;

        END IF;

      END LOOP;

    END LOOP;

  END LOOP;

  dbms_output.put_line('a:' || a_amount);

  dbms_output.put_line('b:' || b_amount);

  dbms_output.put_line('c:' || c_amount);

  dbms_output.put_line('最接近的数字为：' || l_max_amount);

END;