Kann nicht Manager Ausgabe in SQL

Question

Erstellen der Abteilungstabelle

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

Bestücken Abteilung Tabelle

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; 

Erstellen Employee-Tabelle

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)); 

Bestücken Employee-Tabelle

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; 

erhalten

FRAGT: auf den Tabellen DEPARTMENT und EMPLOYEE Basierend einen PL/SQL anonymen Block in Schreib erstellt, die einen Mitarbeiter-ID aus der Benutzereingabe und Anzeigen 1) den Name des Mitarbeiters akzeptiert, Einstellungsdatum und das Gesamtgehalt (Gehalt + Provision), 2) Name des/der Geschäftsführers, Einstellungsdatum und Gesamtentgelt (Gehalt + Provision), 3) Name der Abteilung, in der der Mitarbeiter tätig ist, 4) Anzahl der Mitarbeiter in der Abteilung, in der der Mitarbeiter arbeitet, und 5)

Bis jetzt habe ich die Abfrage unten, ich bin nicht in der Lage, den Namen des Managers, Einstellungsdatum und Gesamtlohn (Gehalt zu erhalten) + commission) Wie kann ich das tun?

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; 

Answer 1

Sie sollten zuerst Mitarbeiter Manager_id erhalten und dann den Manager abfragen. Dies funktioniert für mich:

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; 
/

Und das Ergebnis:

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