m******u
发帖数: 12400 | 1 个人觉得这个题目出的不好（比如第二题），answer key 也不算怎样，第一题就可见
一斑。
。。。。。。。。。。。。。。
Table 1 : DEPT
DEPTNO (NOT NULL , NUMBER(2)), DNAME (VARCHAR2(14)),
LOC (VARCHAR2(13)
Table 2 : EMP
EMPNO (NOT NULL , NUMBER(4)), ENAME (VARCHAR2(10)),
JOB (VARCHAR2(9)), MGR (NUMBER(4)), HIREDATE (DATE),
SAL (NUMBER(7,2)), COMM (NUMBER(7,2)), DEPTNO (NUMBER(2))
MGR is the empno of the employee whom the employee reports to. DEPTNO is a
foreign key.
QUERIES
1. List all the employees who have at least one person reporting to them.
2. List the employee details if and only if more than 10 employees are
present in department no 10.
3. List the name of the employees with their immediate higher authority.
4. List all the employees who do not manage any one.
5. List the employee details whose salary is greater than the lowest salary
of an employee belonging to deptno 20.
6. List the details of the employee earning more than the highest paid
manager.
7. List the highest salary paid for each job.
8. Find the most recently hired employee in each department.
9. In which year did most people join the company? Display the year and the
number of employees.
10. Which department has the highest annual remuneration bill?
11. Write a query to display a ‘*’ against the row of the most recently
hired employee.
12. Write a correlated sub-query to list out the employees who earn more
than the average salary of their department.
13. Find the nth maximum salary.
14. Select the duplicate records (Records, which are inserted, that already
exist) in the EMP table.
15. Write a query to list the length of service of the employees (of the
form n years and m months).
KEYS:
1. SELECT DISTINCT(A.ENAME) FROM EMP A, EMP B WHERE A.EMPNO = B.MGR; or
SELECT ENAME FROM EMP WHERE EMPNO IN (SELECT MGR FROM EMP);
2. SELECT * FROM EMP WHERE DEPTNO IN (SELECT DEPTNO FROM EMP GROUP BY DEPTNO
HAVING COUNT(EMPNO)>10 AND DEPTNO=10);
3. SELECT A.ENAME "EMPLOYEE", B.ENAME "REPORTS TO" FROM EMP A, EMP B WHERE A
.MGR=B.EMPNO;
4. SELECT * FROM EMP WHERE EMPNO IN ( SELECT EMPNO FROM EMP MINUS SELECT MGR
FROM EMP);
5. SELECT * FROM EMP WHERE SAL > ( SELECT MIN(SAL) FROM EMP GROUP BY DEPTNO
HAVING DEPTNO=20);
6. SELECT * FROM EMP WHERE SAL > ( SELECT MAX(SAL) FROM EMP GROUP BY JOB
HAVING JOB = 'MANAGER' );
7. SELECT JOB, MAX(SAL) FROM EMP GROUP BY JOB;
8. SELECT * FROM EMP WHERE (DEPTNO, HIREDATE) IN (SELECT DEPTNO, MAX(
HIREDATE) FROM EMP GROUP BY DEPTNO);
9. SELECT TO_CHAR(HIREDATE,'YYYY') "YEAR", COUNT(EMPNO) "NO. OF EMPLOYEES"
FROM EMP GROUP BY TO_CHAR(HIREDATE,'YYYY') HAVING COUNT(EMPNO) = (SELECT MAX
(COUNT(EMPNO)) FROM EMP GROUP BY TO_CHAR(HIREDATE,'YYYY'));
10. SELECT DEPTNO, LPAD(SUM(12*(SAL+NVL(COMM,0))),15) "COMPENSATION" FROM
EMP GROUP BY DEPTNO HAVING SUM( 12*(SAL+NVL(COMM,0))) = (SELECT MAX(SUM(12*(
SAL+NVL(COMM,0)))) FROM EMP GROUP BY DEPTNO);
11. SELECT ENAME, HIREDATE, LPAD('*',8) "RECENTLY HIRED" FROM EMP WHERE
HIREDATE = (SELECT MAX(HIREDATE) FROM EMP) UNION SELECT ENAME NAME, HIREDATE
, LPAD(' ',15) "RECENTLY HIRED" FROM EMP WHERE HIREDATE != (SELECT MAX(
HIREDATE) FROM EMP);
12. SELECT ENAME,SAL FROM EMP E WHERE SAL > (SELECT AVG(SAL) FROM EMP F
WHERE E.DEPTNO = F.DEPTNO);
13. SELECT ENAME, SAL FROM EMP A WHERE &N = (SELECT COUNT (DISTINCT(SAL))
FROM EMP B WHERE A.SAL<=B.SAL);
14. SELECT * FROM EMP A WHERE A.EMPNO IN (SELECT EMPNO FROM EMP GROUP BY
EMPNO HAVING COUNT(EMPNO)>1) AND A.ROWID!=MIN (ROWID));
15. SELECT ENAME "EMPLOYEE",TO_CHAR(TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/
12))||' YEARS '|| TO_CHAR(TRUNC(MOD(MONTHS_BETWEEN (SYSDATE, HIREDATE),12)))
||' MONTHS ' "LENGTH OF SERVICE" FROM EMP; | x****e
发帖数: 1773 | 2 一眼的*. 这个很不好。
【在 m******u 的大作中提到】
: 个人觉得这个题目出的不好（比如第二题），answer key 也不算怎样，第一题就可见
: 一斑。
: 。。。。。。。。。。。。。。
: Table 1 : DEPT
: DEPTNO (NOT NULL , NUMBER(2)), DNAME (VARCHAR2(14)),
: LOC (VARCHAR2(13)
: Table 2 : EMP
: EMPNO (NOT NULL , NUMBER(4)), ENAME (VARCHAR2(10)),
: JOB (VARCHAR2(9)), MGR (NUMBER(4)), HIREDATE (DATE),
: SAL (NUMBER(7,2)), COMM (NUMBER(7,2)), DEPTNO (NUMBER(2))
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