Correct Answer
verified
Short Answer
Correct Answer
verified
Multiple Choice
A) SELECT title
FROM books, orderitems
WHERE books.isbn <> orderitems.isbn;
B) SELECT title
FROM books NATURAL JOIN orderitems
WHERE JOIN IS NULL;
C) SELECT title FROM books NATURAL JOIN orderitems
MINUS
SELECT title FROM books;
D) SELECT title FROM books
MINUS
SELECT title FROM books NATURAL JOIN orderitems;
Correct Answer
verified
Multiple Choice
A) outer join
B) non-equality join
C) self-join
D) Cartesian join
Correct Answer
verified
True/False
Correct Answer
verified
Multiple Choice
A) SELECT customer#, lastname, firstname, order#
FROM customers NATURAL JOIN orders;
B) SELECT customer#, lastname, firstname, order#
FROM customers JOIN orders USING (customer#) ;
C) SELECT c.customer#, lastname, firstname, order# FROM customers c, orders o;
D) both a and b
Correct Answer
verified
True/False
Correct Answer
verified
True/False
Correct Answer
verified
True/False
Correct Answer
verified
Multiple Choice
A) SELECT lastname, firstname, order#
FROM customers NATURAL JOIN orders
WHERE orders.customer# IS NOT NULL;
B) SELECT lastname, firstname, order#
FROM customers, orders
WHERE orders.customer# (+) = customers.customer#;
C) SELECT lastname, firstname, order#
FROM customers, orders
WHERE orders.customer# = customers.customer# (+) ;
D) SELECT lastname, firstname, order#
FROM customers NATURAL JOIN orders
WHERE orders.customer# IS NULL;
Correct Answer
verified
True/False
Correct Answer
verified
Short Answer
Correct Answer
verified
Short Answer
Correct Answer
verified
Multiple Choice
A) SELECT title
FROM books NATURAL JOIN orderitems
WHERE qty > 1;
B) SELECT title
FROM books JOIN orderitems
WHERE qty > 1;
C) SELECT title
FROM books JOIN orderitems ON (isbn) JOIN orders ON (order#)
WHERE qty>1;
D) SELECT title
FROM books JOIN orderitems USING(isbn) ;
Correct Answer
verified
Multiple Choice
A) SELECT r.firstname, r.lastname, c.customer# FROM customers r, customers c
WHERE r.customer# = c.referred;
B) SELECT r.firstname, r.lastname, c.customer# FROM customers r JOIN customers c
ON r.customer# = c.referred;
C) SELECT r.firstname, r.lastname, c.customer# FROM customers r NATURAL JOIN customers c;
D) both a and b
Correct Answer
verified
True/False
Correct Answer
verified
True/False
Correct Answer
verified
Multiple Choice
A) comma (,)
B) plus sign (+)
C) period (.)
D) percent sign (%)
Correct Answer
verified
Multiple Choice
A) FULL OUTER JOIN
B) JOIN...ON
C) NATURAL JOIN
D) CROSS JOIN
Correct Answer
verified
True/False
Correct Answer
verified
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