Oracle 在表中找不到给定记录时，如何处理游标中的异常

我正在使用循环游标更新

emp3

（与emp表相同）

DECLARE 
   CURSOR incr_cur IS SELECT * FROM emp3 FOR UPDATE OF sal;
   v_job emp3.job%TYPE := '&ENTER_Job';
   v_cnt INTEGER := 0;
BEGIN
  FOR r_l IN incr_cur LOOP
    IF v_job IN (r_l.job) THEN
      UPDATE emp3 SET sal = sal + 100 WHERE CURRENT OF incr_cur;
    END IF;
 END LOOP;

 FOR r_l IN incr_cur LOOP
    IF v_job IN (r_l.job) THEN
      v_cnt := v_cnt + 1;
      DBMS_OUTPUT.PUT_LINE('The Salary of ' || r_l.ename || ' is Incremented by 100 and the Updated Salary is: $' || r_l.sal);
    END IF;
 END LOOP;

     DBMS_OUTPUT.PUT_LINE('The Salary of '|| v_cnt ||' Employees are Updated');
END;

执行PL/SQL块时，它将请求作业， 我给

经理

，然后担任

经理

的员工的工资增加100。

emp3

表有5个工作类别

职员

，

经理

，

分析师

，

推销员

和

总裁

。 然后如何显示消息作业未列出，因此无法进行更新。，如果用户输入的作业不在表中，例如

DEVELOPER

。

---

我尝试过异常处理，但无法让它工作。

这里有一个选项：检查是否存在这样的作业；如果没有，查询将返回

未找到任何数据

，您可以处理该数据并使用适当的消息引发异常。否则，继续执行

更新

SQL> declare
  2    l_job emp.job%type;
  3  begin
  4    begin
  5      select job
  6        into l_job
  7        from emp
  8        where job = '&ENTER_Job'
  9          and rownum = 1;
 10    exception
 11      when no_data_found then
 12        raise_application_error(-20000, 'That job does not exist');
 13    end;
 14
 15    -- Job exists, so - go on with the update
 16  end;
 17  /
Enter value for enter_job: MANAGER

PL/SQL procedure successfully completed.

SQL> /
Enter value for enter_job: DEVELOPER
declare
*
ERROR at line 1:
ORA-20000: That job does not exist
ORA-06512: at line 12


SQL>

另外，我忘了提到：我更喜欢通过存储过程（接受作业名称作为参数）而不是匿名PL/SQL块来执行此类作业。
不需要单独的步骤。只要尝试更新，如果没有行被更新，就说出来。如果希望它是一个异常，则使用
raise\u application\u error
引发一个异常

假设这是一个学习练习，这就是为什么您不想只做一个普通的
更新
，您可以这样做：

declare 
    k_job constant emp3.job%type := '&JOB';

    cursor employees_cur is
        select * from emp3
        where  job = k_job
        for update of sal;

    v_update_count integer := 0;
    v_payroll_increase integer := 0;
begin
    for r in employees_cur loop
        update emp3 set sal = sal + 100 where current of employees_cur;
        dbms_output.put_line('Salary for ' || r.ename || ' is incremented by $100 from $' || r.sal || ' to $' || (r.sal +100));
        v_update_count := v_update_count + 1;
        v_payroll_increase := v_payroll_increase + 100;
    end loop;

    if v_update_count = 0 then
        dbms_output.put_line('No staff are currently employed as ' || k_job ||'. Payroll is unchanged.');
    else
        dbms_output.put_line('Updated salary of '|| v_update_count ||' employee' || case when v_update_count <> 1 then 's' end||'.');
        dbms_output.put_line('Payroll increased by $'||v_payroll_increase||'.');
    end if;
end;
/

Enter value for job: SALESMAN
Salary for ALLEN is incremented by $100 from $1600 to $1700
Salary for WARD is incremented by $100 from $1250 to $1350
Salary for MARTIN is incremented by $100 from $1250 to $1350
Salary for TURNER is incremented by $100 from $1500 to $1600
Updated salary of 4 employees.
Payroll increased by $400.

PL/SQL procedure successfully completed.

（在本例中，
v_payroll\u increase
总是100倍
v_update\u count
，但如果您想按部门等给出10%的加薪或不同的加薪，可能会更有用。）
您为什么不使用带有
where
子句的单个SQL
update
语句来完成整个过程？
Enter value for job: ASTRONAUT
No staff are currently employed as ASTRONAUT. Payroll is unchanged.