Database Iitu Quiz

True

False

&

-

%

*

SELECT last_name Name FROM employees WHERE Name like ‘K%’

SELECT department_id Dep, COUNT(last_name) FROM employees GROUP BY dep

SELECT department_id Dep, COUNT(last_name) Num FROM employees GROUP BY department_id HAVING Num>5

SELECT last_name Name FROM employees ORDER BY Name

DELETE FROM zipcode WHERE zip ANY (‘02199’,’43011’) COMMIT

DELETE FROM zipcode WHERE zip ALL (‘02199’,’43011’) COMMIT

DELETE FROM zipcode WHERE zip IN (‘02199’,’43011’) COMMIT

DELETE FROM zipcode WHERE zip ANY (‘02199’,’43011’)

Select ename from emp where hiredate < add_month(sysdate,-60);

Select ename from emp where hiredate < add_month(sysdate,+5);

Select ename from emp where hiredate < add_month(sysdate,+60);

Select ename from emp where hiredate < add_month(sysdate,-5);

An error will occur since none of the JOIN operations is used

The Cartesian product of the two tables based on the department_id column will be displayed

The last names of the employees and their salaries will be displayed along with the departments names where the employees work

An error will occur since in the SELECT clause a column thet is used to join the tables is missing

The keyboard AS between an alias and a column name is optional

Aliases always require double quotation marks

Aliases immediately follows the column name

Aliases rename column heading

WHERE salary > 5000 OR salary < 18000

WHERE salary < 5000 AND salary > 18000

WHERE salary > 5000 AND salary < 18000

WHERE salary >5000 AND salary > 18000

ORDER BY salary;

ORDER BY 2

ORDER BY “Salary” ASC;

ORDER BY “Salary” DESC;

WHERE

ORDER BY

FROM

GROUP BY

CONCAT

SUM

UPPER

ROUND

SELECT department_id, COUNT(last_name) FROM employees GROUP BY department_id

SELECT department_name, COUNT(last_name) FROM employees GROUP BY department_id

SELECT department_id, department_name, COUNT(last_name) FROM employees GROUP BY department_id

SELECT department_id, COUNT(last_name) FROM departments GROUP_BY last_name

True

False

SELECT * FROM employees WHERE last_name=’&Name’

SELECT job_id, SUM(salary) FROM employees GROUP BY ‘&Column’

SELECT * FROM &Table

SELECT &Column FROM employees

Select deptno, count(deptno) from emp group by deptno having count(*)>3;

Select deptno, count (deptno) from emp group by deptno having count(*)>2;

Select deptno, count (emp) from deptno group by emp having count(*)>2;

Select deptno, count(deptno) from deptno group by emp having count(*)>3;

Compound queries

Subqueries

Inner queries

Outer queries

True

False

Multiple-row subqueries

Multiple-column subqueries

Sing-column subqueries

Single-row subqueries

UNION ALL

MINUS

INTERSECT

UNION

HAVING

ORDER BY

GROUP BY

WHERE

True

False

SAVEPOINT

ROLLBACK

COMMIT

DISCARD

SELECT * FROM userb.employees;

SELECT * FROM employees;

SELECT * FROM usera.employees;

SELECT * FROM employees.userb;

Main query

Subquery

Inner query

Outer query

UNION ALL

INTERSECT

UNION

MINUS

An error will occur, since SUBSTR function need to take 3 parameters

Hello,World!!!13

!!!

Hello, World!

WHERE salary>5000 AND job_id = ‘IT_PROG’

WHERE salary>5000 AND job_id = “IT_PROG”

WHERE salary>5000 AND job_id = IT_PROG

WHERE salary>5000 OR job_id = ‘IT_PROG’

-

+

%

*

WHERE last_name = ‘K%’ and first_name = ‘K%’

WHERE last_name LIKE ‘K_’ and first_name LIKE ‘K_’

WHERE last_name LIKE ‘K%’ and first_name LIKE ‘K%’

WHERE last_name and first_name = ‘K%’

SELECT… WHERE … ORDER BY … GROUP BY

SELECT … FROM … GROUP BY … ORDER BY … WHERE

SELECT…FROM…WHERE…GROUP BY…ORDER BY

SELECT … FROM … WHERE … ORDER BY … GROUP BY

Update emp set sal=sal*1.1 where empno in (select mgr from emo);

Update emp set sal=sal*1.1 where empno in (select emp from emo);

Update emp set sal=sal*0.1 where empno in (select emp from emo);

Update emp set sal=sal*0.1 where empno in (select mgr from emo);

64

60

63

70

Double quotes

Single quotes

Parantheses

Braces

&

&

||

*

First, outer

Last, outer

Last, inner

First, inner

True

False