.
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
True
False
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);
True
False
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
True
False
SELECT * FROM userb.employees;
SELECT * FROM employees;
SELECT * FROM usera.employees;
SELECT * FROM employees.userb;
Main query
Subquery
Inner query
Outer query
True
False
UNION ALL
INTERSECT
UNION
MINUS
True
False
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’
True
False
-
+
%
*
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);
True
False
64
60
63
70
True
False
True
False
True
False
Double quotes
Single quotes
Parantheses
Braces
&
&
||
*
First, outer
Last, outer
Last, inner
First, inner
True
False
True
False
True
False
True
False
True
False
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