首先,我得到的错误(在创建employees
表之后)不仅仅是“错误(8,5):PL/SQL:语句被忽略”,而是
LINE/COL ERROR
---------- ---------------------------------------------------------
6/5 PL/SQL: Statement ignored
6/36 PLS-00201: identifier 'FIRST_NAME' must be declared
9/5 PL/SQL: SQL Statement ignored
9/32 PLS-00201: identifier 'PASSWORD' must be declared
9/41 PL/SQL: ORA-00904: : invalid identifier
因此,了解如何在您使用的任何工具中报告编译错误是值得的。
错误告诉您FIRST_NAME
并且PASSWORD
未声明为变量。
一旦你声明了缺失的变量,你需要使用一个select into
构造来填充它们,指定employee_id。就个人而言,我会将整个员工行提取到记录中,而不是管理多个单独的变量。
create or replace procedure set_password(e_id number)
is
emp employees%rowtype;
begin
select * into emp
from employees e
where e.employee_id = e_id;
emp.password :=
lower(substr(emp.first_name, 1, 1)) ||
lower(emp.last_name) ||
emp.employee_id ||
substr(emp.phone_number, 1, 3);
update employees e
set password = emp.password
where employee_id = e_id;
dbms_output.put_line('El password para el empleado ' || e_id || ' se ha generado exitosamente');
dbms_output.put_line('El nuevo password es: ' || emp.password);
end set_password;
测试值:
create table employees
( employee_id
, first_name
, last_name
, phone_number
, password )
as
select 123
, 'Karl'
, 'Marx'
, '07345678910'
, cast('' as varchar2(50))
from dual;
call set_password(123);
El password para el empleado 123 se ha generado exitosamente
El nuevo password es: kmarx123073
从安全角度来看,您确定要生成可预测的密码吗?如果是我,我宁愿将其设置为dbms_random.string('a',10)
.