如何在表中找不到给定记录时如何处理游标中的异常

时间:2018-03-31 07:20:03

标签: oracle plsql oracle10g

我正在使用Cursor For循环更新loop.call_soon_threadsafe(event.set)(与emp表相同)

emp3

执行PL / SQL块时,它会询问作业, 我给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; ,然后MANAGER的员工的工资增加100。 MANAGER表有5个JOB类别emp3CLERKMANAGERANALYSTSALESMAN。 然后如何显示消息未列出作业,因此无法更新。,如果用户输入了PRESIDENT表中不存在的作业。 我曾尝试过异常处理,但无法让它工作。

2 个答案:

答案 0 :(得分:1)

这里有一个选择:检查这样的工作是否存在;如果没有,查询将返回您可以处理的NO_DATA_FOUND,并使用适当的消息引发异常。否则,请继续UPDATE

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>

P.S。忘记提及: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块来完成这样的工作。

答案 1 :(得分:1)

不需要单独的步骤。只是尝试更新,如果没有更新行,请说明。如果您希望它成为例外,请使用raise_application_error举一个。

假设这是一个学习练习,这就是为什么你不想只做一个普通的update,你可能会做这样的事情:

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.

对于不存在的工作,你得到这个:

Enter value for job: ASTRONAUT
No staff are currently employed as ASTRONAUT. Payroll is unchanged.

(在此示例中,v_payroll_increase始终为v_update_count的100倍,但如果您希望按部门等提高10%或增加不同,则可能更有用。)