程序员开发人员试题

时间:2025-07-10

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

1、哪一个SQL有错误?

A. SELECT * FROM emp_dept_vu;

B. SELECT department_id, SUM(salary)

FROM emp_dept_vu

GROUP BY department_id;

C. SELECT department_id, job_id, AVG(salary)

FROM emp_dept_vu

GROUP BY department_id, job_id;

D. SELECT job_id, SUM(salary)

FROM emp_dept_vu

WHERE department_id IN (10,20)

GROUP BY job_id

HAVING SUM(salary) > 20000;

E. 没有一个错误,都是对的.

2、两张表结构如下:

EMPLOYEES

(EMPLOYEE_ID NUMBER Primary Key

FIRST_NAME VARCHAR2(25)

LAST_NAME VARCHAR2(25)

HIRE_DATE DATE

NEW_EMPLOYEES

EMPLOYEE_ID NUMBER Primary Key

NAME VARCHAR2(60)

哪一个删除语句是正确的。

A. DELETE FROM employees

WHERE employee_id = (SELECT employee_id

FROM employees);

B. DELETE * FROM employees

WHERE employee_id = (SELECT employee_id

FROM new_employees);

C. DELETE FROM employees

WHERE employee_id IN (SELECT employee_id

FROM new_employees

WHERE name ='Carrey');

D. DELETE * FROM employees

WHERE employee_id IN (SELECT employee_id

FROM new_employees

WHERE last_name ='Carrey');

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

3、显示每个部门,最高SALARY,最小SALARY,且最小SALARY小于5000,最大SALARY大于15000。

A. SELECT dept_id, MIN(salary), MAX(salary)

FROM employees

WHERE MIN(salary) < 5000 AND MAX(salary) > 15000;

B. SELECT dept_id, MIN(salary), MAX(salary)

FROM employees

WHERE MIN(salary) < 5000 AND MAX(salary) > 15000

GROUP BY dept_id;

C. SELECT dept_id, MIN(salary), MAX(salary)

FROM employees

HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;

D. SELECT dept_id, MIN(salary), MAX(salary)

FROM employees

GROUP BY dept_id

HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;

E. SELECT dept_id, MIN(salary), MAX(salary)

FROM employees

GROUP BY dept_id, salary

HAVING MIN(salary) < 5000 AND MAX(salary) > 15000;

4、执行SQL:

SELECT * FROM orders

WHERE cust_id = (SELECT cust_id FROM customers WHERE cust_name = 'Smith');

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

5、students表结构如下:

STD_ID NUMBER(4)

COURSE_ID VARCHAR2(10)

START_DATE DATE

END_DATE DATE

哪两个在列START_DATE上的运算是正确的?

A. SUM(start_date)

B. AVG(start_date)

C. COUNT(start_date)

D. AVG(start_date, end_date)

E. MIN(start_date)

F. MAXIMUM(start_date)

6、EMPLOYEES表结构如下:

EMP_ID NUMBER(4) NOT NULL

LAST_NAME VARCHAR2(30) NOT NULL

FIRST_NAME VARCHAR2(30)

DEPT_ID NUMBER(2)

哪个SQL语句实现last_name 为 ‘Smith’不同DEPT_ID的总数?

A. SELECT COUNT(*)

FROM employees

WHERE last_name='Smith';

B. SELECT COUNT(dept_id)

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

FROM employees

WHERE last_name='Smith';

C. SELECT DISTINCT(COUNT(dept_id))

FROM employees

WHERE last_name='Smith';

D. SELECT COUNT(DISTINCT dept_id)

FROM employees

WHERE last_name='Smith';

E. SELECT UNIQUE(dept_id)

FROM employees

WHERE last_name='Smith';

7、查询所有employees表中的数据,无论在departments表中是否department_id的记录。哪个查询是正确的?

A. SELECT last_name, department_name

FROM employees NATURAL JOIN departments;

B. SELECT last_name, department_name

FROM employees JOIN departments ;

C. SELECT last_name, department_name

FROM employees e JOIN departments d

ON (e.department_id = d.department_id);

D. SELECT last_name, department_name

FROM employees e

RIGHT OUTER JOIN departments d ON (e.department_id = d.department_id); E. SELECT last_name, department_name

FROM employees FULL JOIN departments

ON (e.department_id = d.department_id);

F. SELECT last_name, department_name

FROM employees e LEFT OUTER

JOIN departments d ON (e.department_id = d.department_id);

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

8、执行以下SQL,哪个结果是正确的。

SELECT e.employee_id "Emp_id", e.emp_name "Employee", e.salary,

m.employee_id "Mgr_id", m.emp_name "Manager"

FROM employees e JOIN employees m

ON (e.mgr_id = m.employee_id)

AND e.salary > 4000;

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

9、N张表等值联合查询,至少要写几个连接条件?

A. N

B. N-1

C. N+1

10、哪三个查询没有问题?

程序员开发人员面试试题.包括linux基本操作和sql的一些基本操作.

A. SELECT *

FROM employees

where salary > (SELECT MIN(salary)

FROM employees

GROUP BY department_id);

B. SELECT *

FROM employees

WHERE salary = (SELECT AVG(salary)

FROM employees

GROUP BY department_id);

C. SELECT distinct department_id

FROM employees

WHERE salary > ANY (SELECT AVG(salary)

FROM employees

GROUP BY department_id);

D. SELECT department_id

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