A. ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
B. ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
C. ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;
D. ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;
E. ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;
F. ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;

A. The sort is in ascending by order by default.
B. The sort is in descending order by default.
C. The ORDER BY clause must precede the WHERE clause.
D. The ORDER BY clause is executed on the client side.
E. The ORDER BY clause comes last in the SELECT statement.
F. The ORDER BY clause is executed first in the query execution.

A. Convert 10 to ‘TEN’
B. Convert ‘10’ to 10
C. Convert ‘10’ to ‘10’
D. Convert ‘TEN’ to 10
E. Convert a date to a character expression
F. Convert a character expression to a date

A. the use of rowid
B. a GROUP BY clause
C. an ORDER BY clause
D. only an inline view
E. an inline view and an outer query

A. SELECT TOTAL(*)   FROM customers;
B. SELECT COUNT(*)   FROM customers;
C. SELECT TOTAL(customer_id)   FROM customers;
D. SELECT COUNT(customer_id)   FROM customers;
E. SELECT COUNT(customers)   FROM customers;
F. SELECT TOTAL(customer_name)   FROM customers;

A. SELECT COUNT (*) FROM employees WHERE last _name='smith';
B. SELECT COUNT (dept_id) 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 UNIQE (dept_id) FROM employees WHERE last _name='smith';

A. INSERT
B. UPDATE
C. SELECT
D. DESCRIBE
E. DELETE
F. RENAME

A. DROP emp_dept_uv;
B. DELETE emp_dept_uv;
C. REMOVE emp_dept_uv;
D. DROP VIEW emp_dept_uv;
E. DELETE VIEW emp_dept_uv;
F. REMOVE VIEW emp_dept_uv;

A. Deleting the records of employees who do not earn commission.
B. Increasing the commission of employee 3 by the average commission earned in department 20.
C. Finding the number of employees who do NOT earn commission and are working for department 20.
D. Inserting into the table a new employee 10 who works for department 20 and earns a commission that is equal to the commission earned by employee 3.
E. Creating a table called COMMISSION that has the same structure and data as the columns EMP_ID and COMMISSIONS of the EMP table.
F. Decreasing the commission by 150 for the employees who are working in department 30 and earning a commission of more then 800.

A. CASCADE
B. UNIQUE
C. NONUNIQUE
D. CHECK
E. PRIMARY KEY
F. CONSTANT
G. NOT NULL

A. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, 'a') FROM EMPLOYEES WHERE SUBSTR (ENAME, -1,1) = 'n';
B. SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, ,-1,1) FROM EMPLOYEES WHERE SUBSTR (ENAME, -1,1) = 'n';
C. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR (ENAME, 1,1) = 'n';
D. SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR (ENAME, -1,1) = 'n';