无法在 SQL 中获取管理器输出答案

【问题标题】：Can't get manager output in SQL无法在 SQL 中获取管理器输出
【发布时间】：2016-04-12 08:13:21
【问题描述】：

创建部门表

CREATE TABLE department
( DEPARTMENT_ID         NUMBER(4)       PRIMARY KEY,    
  DEPARTMENT_NAME   VARCHAR2(20)    NOT NULL UNIQUE,
  ADDRESS           VARCHAR2(20)    NOT NULL);

填充部门表

INSERT INTO department VALUES(10, 'ACCOUNTING', 'NEW YORK');
INSERT INTO department VALUES(20, 'RESEARCH', 'DALLAS');
INSERT INTO department VALUES(30, 'SALES', 'CHICAGO');
INSERT INTO department VALUES(40, 'IT', 'DALLAS');
INSERT INTO department VALUES(50, 'EXECUTIVE', 'NEW YORK');
INSERT INTO department VALUES(60, 'MARKETING', 'CHICAGO');
COMMIT;

创建员工表

CREATE TABLE employee 
( EMPLOYEE_ID   NUMBER(4)       PRIMARY KEY,
  EMPLOYEE_NAME VARCHAR2(20)    NOT NULL,
  JOB           VARCHAR2(50)    NOT NULL,
  MANAGER_ID    NUMBER(4),
  HIRE_DATE     DATE            NOT NULL,
  SALARY        NUMBER(9, 2)    NOT NULL,
  COMMISSION    NUMBER(9, 2),
  DEPARTMENT_ID NUMBER(4) REFERENCES department(DEPARTMENT_ID));

填充员工表

INSERT INTO employee 
VALUES(7839, 'KING',  'PRESIDENT', NULL, '20-NOV-01', 5000, NULL, 50);
INSERT INTO employee 
VALUES(7596, 'JOST',  'VICE PRESIDENT', 7839, '04-MAY-01', 4500, NULL, 50);
INSERT INTO employee  
VALUES(7603, 'CLARK', 'VICE PRESIDENT', 7839, '12-JUN-01', 4000, NULL, 50);
INSERT INTO employee 
VALUES(7566, 'JONES', 'PUBLIC ACCOUNTANT', 7596, '05-APR-01', 3000, NULL,     10);
INSERT INTO employee 
VALUES(7886, 'STEEL', 'PUBLIC ACCOUNTANT', 7566, '08-MAR-03', 2500, NULL, 10);
INSERT INTO employee 
VALUES(7610, 'WILSON', 'ANALYST', 7596, '03-DEC-01', 3000, NULL, 20);
INSERT INTO employee  
VALUES(7999, 'WOLFE',  'ANALYST', 7610, '15-FEB-02', 2500, NULL, 20);
INSERT INTO employee  
VALUES(7944, 'LEE', 'ANALYST', 7610, '04-SEP-06', 2400, NULL, 20);
INSERT INTO employee 
VALUES(7900, 'FISHER', 'SALESMAN', 7603, '06-DEC-01', 3000, 500, 30);
INSERT INTO employee 
VALUES(7921, 'JACKSON', 'SALESMAN', 7900, '25-FEB-05', 2500, 400, 30);
INSERT INTO employee 
VALUES(7952, 'LANCASTER', 'SALESMAN',  7900, '06-DEC-06', 2000, 150, 30);
INSERT INTO employee  
VALUES(7910, 'SMITH', 'DATABASE ADMINISTRATOR', 7596, '20-DEC-01', 2900,     NULL, 40);
INSERT INTO employee  
VALUES(7788, 'SCOTT', 'PROGRAMMER', 7910, '15-JAN-03', 2500, NULL, 40);
INSERT INTO employee 
VALUES(7876, 'ADAMS', 'PROGRAMMER', 7910, '15-JAN-03', 2000, NULL, 40);
INSERT INTO employee 
VALUES(7934, 'MILLER','PROGRAMMER', 7876, '25-JAN-02', 1000, NULL, 40);
INSERT INTO employee 
VALUES(8000, 'BREWSTER',  'TBA',  NULL,   '22-AUG-13', 2500, NULL, NULL);
COMMIT;

问题：基于在编写 PL/SQL 匿名块中创建的 DEPARTMENT 和 EMPLOYEE 表，该块接受来自用户输入的员工 ID 并显示 1) 员工姓名、雇用日期和总数工资（工资+佣金），2）他/她的经理姓名、入职日期和总工资（工资+佣金），3）员工所在部门的名称，4）所在部门的员工人数员工工作，5）员工所在部门的总工资（工资+佣金）之和，

到目前为止，我无法获得经理的姓名、雇用日期和总工资（薪水 + 佣金），我该怎么做？

ACCEPT p_1 PROMPT 'Please enter the employee id:'

DECLARE
v_input_emp_id NUMBER := &p_1;
v_emp_name employee.employee_name%TYPE;
v_dept_name department.department_name%TYPE;
v_pay employee.salary%TYPE;
v_n NUMBER;
v_output VARCHAR2(100) := 'The employee id is not in the table';
v_hiredate EMPLOYEE.HIRE_DATE%TYPE;

BEGIN
SELECT COUNT(*)
INTO v_n
FROM employee
WHERE employee_id = v_input_emp_id;

IF v_n = 0 THEN
DBMS_OUTPUT.PUT_LINE(v_input_emp_id || ' is not in the EMPLOYEE table.');
END IF;

IF v_n = 1 THEN

SELECT employee_name, salary + NVL(commission, 0),HIRE_DATE
INTO v_emp_name, v_pay, v_hiredate
FROM employee
WHERE employee_id = v_input_emp_id;

DBMS_OUTPUT.PUT_LINE('USER INPUT:');
DBMS_OUTPUT.PUT_LINE(v_input_emp_id);

DBMS_OUTPUT.PUT_LINE('----------EMPLOYEE----------');
DBMS_OUTPUT.PUT_LINE(rpad('NAME:',10,' ') || lpad(v_emp_name,15));
DBMS_OUTPUT.PUT_LINE(rpad('HIRE DATE:',10) || lpad(v_hiredate,20));
DBMS_OUTPUT.PUT_LINE('TOTAL PAY:' || lpad(TO_CHAR(ROUND(v_pay, 2),
'$9,990.00'),21));

SELECT COUNT(*)
INTO v_n
FROM employee
WHERE manager_id = v_input_emp_id;

IF v_n != 0 THEN

SELECT employee_name ,salary + NVL(commission, 0),HIRE_DATE
INTO v_emp_name , v_pay, v_hiredate 
FROM employee
WHERE employee.manager_id = v_input_emp_id;

DBMS_OUTPUT.PUT_LINE('----------MANAGER----------');
DBMS_OUTPUT.PUT_LINE(rpad('NAME:',10,' ') || lpad(v_emp_name,15));
DBMS_OUTPUT.PUT_LINE(rpad('HIRE DATE:',10) || lpad(v_hiredate,20));
DBMS_OUTPUT.PUT_LINE('TOTAL PAY:' || lpad(TO_CHAR(ROUND(v_pay, 2),
'$9,990.00'),21));

END IF;
DBMS_OUTPUT.PUT_LINE('----------MANAGER----------');
DBMS_OUTPUT.PUT_LINE(rpad('NAME:',10,' ') || lpad(v_emp_name,15));

ELSE 
DBMS_OUTPUT.PUT_LINE('EMPLOYEE NOT IN THE TABLE');

END IF;
END;

【问题讨论】：

经理也应该是员工，因此尝试将员工经理的 id 选择到 v_manager_id 中。然后从员工那里查询employee_id = v_manager_id;
刚刚试了一下，它不起作用我没有得到任何输出

标签： sql database oracle11g

【解决方案1】：

你应该先获取员工 manager_id，然后查询经理。这对我有用：

SET SERVEROUTPUT ON
ACCEPT p_1 PROMPT 'Please enter the employee id:'

DECLARE
  v_input_emp_id NUMBER := &p_1;
  v_emp_name     employee.employee_name%TYPE;
  v_dept_name    department.department_name%TYPE;
  v_pay          employee.salary%TYPE;
  v_mangager_id  employee.manager_id%TYPE;
  v_n            NUMBER;
  v_output       VARCHAR2(100) := 'The employee id is not in the table';
  v_hiredate     employee.hire_date%TYPE;

BEGIN
  SELECT COUNT(*) INTO v_n FROM employee WHERE employee_id = v_input_emp_id;

  IF v_n = 0 THEN
    dbms_output.put_line(v_input_emp_id || ' is not in the EMPLOYEE table.');
  END IF;

  IF v_n = 1 THEN

    SELECT employee_name
          ,salary + nvl(commission, 0)
          ,hire_date
          ,manager_id
      INTO v_emp_name
          ,v_pay
          ,v_hiredate
          ,v_mangager_id
      FROM employee
     WHERE employee_id = v_input_emp_id;

    dbms_output.put_line('USER INPUT:');
    dbms_output.put_line(v_input_emp_id);

    dbms_output.put_line('----------EMPLOYEE----------');
    dbms_output.put_line(rpad('NAME:', 10, ' ') || lpad(v_emp_name, 15));
    dbms_output.put_line(rpad('HIRE DATE:', 10) || lpad(v_hiredate, 20));
    dbms_output.put_line('TOTAL PAY:' || lpad(to_char(round(v_pay, 2), '$9,990.00'), 21));

    SELECT COUNT(*) INTO v_n FROM employee WHERE employee_id = v_mangager_id;

    IF v_n != 0 THEN

      SELECT employee_name
            ,salary + nvl(commission, 0)
            ,hire_date
        INTO v_emp_name
            ,v_pay
            ,v_hiredate
        FROM employee
       WHERE employee.employee_id = v_mangager_id;

      dbms_output.put_line('----------MANAGER----------');
      dbms_output.put_line(rpad('NAME:', 10, ' ') || lpad(v_emp_name, 15));
      dbms_output.put_line(rpad('HIRE DATE:', 10) || lpad(v_hiredate, 20));
      dbms_output.put_line('TOTAL PAY:' || lpad(to_char(round(v_pay, 2), '$9,990.00'), 21));

    END IF;
    dbms_output.put_line('----------MANAGER----------');
    dbms_output.put_line(rpad('NAME:', 10, ' ') || lpad(v_emp_name, 15));

  ELSE
    dbms_output.put_line('EMPLOYEE NOT IN THE TABLE');

  END IF;
END;
/

结果：

USER INPUT:
7610
----------EMPLOYEE----------
NAME:              WILSON
HIRE DATE:          2001.12.03
TOTAL PAY:            $3,000.00
----------MANAGER----------
NAME:                JOST
HIRE DATE:          2001.05.04
TOTAL PAY:            $4,500.00
----------MANAGER----------
NAME:                JOST
PL/SQL procedure successfully completed

【讨论】：