Plsql SQL过程问题

Plsql SQL过程问题,plsql,oracle18c,Plsql,Oracle18c,我目前有两个表要处理(如下所列)。下面是我当前程序的代码,它接受employeeID值并输出其信息。此外,我需要对其进行修改,以列出谁是部门主席,如果该员工已经是部门负责人,则将其列为“部门主席” 我尝试创建一个单独的视图来告诉谁是系主任,并循环调用,但没有成功。我被困在这里了 create or replace procedure emp_records ( p_emp_id employees.employeeid%type ) is l_department_name

我目前有两个表要处理(如下所列)。下面是我当前程序的代码,它接受employeeID值并输出其信息。此外,我需要对其进行修改,以列出谁是部门主席,如果该员工已经是部门负责人,则将其列为“部门主席”

我尝试创建一个单独的视图来告诉谁是系主任,并循环调用,但没有成功。我被困在这里了

create or replace procedure emp_records (
    p_emp_id employees.employeeid%type
) is

    l_department_name        departments.departmentname%type;
    l_department_id          departments.departmentid%type;
    l_employee_first_name    employees.firstname%type;
    l_employee_last_name     employees.lastname%type;
    l_employee_employee_id   employees.employeeid%type;
    l_chair_id               departments.departmentchair%type;
    l_chair_name             varchar2(100);
begin
    -- For employee details
    select
        departmentname,
        firstname,
        lastname,
        employeeid,
        departmentid
    into
        l_department_name,
        l_employee_first_name,
        l_employee_last_name,
        l_employee_employee_id,
        l_department_id
    from departments d
    join employees e 
      on e.departmentid = d.departmentidusing
   where employeeid = p_emp_id;

   --For CHair details
  select departmentchair
    into l_chair_id
    from department
   where departmentid = l_department_id;

    if l_chair_id = l_employee_employee_id 
        then
            select 'Is the department chair' into l_chair_name from dual;
    else
        select 'Department Chair: ' || firstname || ' ' || lastname
          into l_chair_name
          from employees
         where employeeid = l_chair_id;
    end if;

    -- For output
    dbms_output.put_line('Information for Employee #' || p_emp_id);
    dbms_output.put_line('Full Name: ' || l_employee_first_name || ' ' || l_employee_last_name);
    dbms_output.put_line('Department: ' || l_department_name);
    dbms_output.put_line(l_chair_name);
exception
    when no_data_found then
        dbms_output.put_line('No records found for this Employee.');
end;
/

希望这会有所帮助。

为了提供解决方案,我将复制您的代码并修复它。在发布之前,我需要测试我为您提供的任何解决方案。为了测试它,我需要样本数据。请向我解释如何使用示例数据创建和填充示例数据库表,这些数据来自您上传的图像,而无需自己编写SQL。如果你不知道,我建议你读一读