Book Sales Database: Which of the following lists all books published by the publisher named Printing Is Us? SELECT title FROM books NATURAL JOIN publisher WHERE name = PRINTING IS US'; SELECT title FROM books, publisher WHERE pubname =1; SELECT title FROM books, publisher WHERE pubname = 1; %3D none of above

Oracle 12c: SQL
3rd Edition
ISBN:9781305251038
Author:Joan Casteel
Publisher:Joan Casteel
Chapter2: Basic Sql Select Statements
Section: Chapter Questions
Problem 10RQ: What is a NULL value?
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Book Sales Database: Which of the following lists all books published by the
publisher named Printing Is Us?
SELECT title
FROM books NATURAL JOIN publisher
WHERE name = 'PRINTING IS US';
SELECT title
FROM books, publisher
WHERE pubname =1;
SELECT title
FROM books, publisher
WHERE pubname = 1;
none of above
Transcribed Image Text:Book Sales Database: Which of the following lists all books published by the publisher named Printing Is Us? SELECT title FROM books NATURAL JOIN publisher WHERE name = 'PRINTING IS US'; SELECT title FROM books, publisher WHERE pubname =1; SELECT title FROM books, publisher WHERE pubname = 1; none of above
Book Sales database (Just Lee Book sales Company): Which of below SQL option
resolves this problem?
Produce a list of each customer number and the third and fourth digits of his or her
zip
code. The query should also display the position of the first occurrence of a 3 in the
customer number, if it exists.
SELECT customer#, SUBSTR(zip, 3), INSTR(customer#, 3)
FROM customers;
SELECT customer#, INSTR(zip, 3, 2), INSTR(customer#, 3)
FROM customers;
SELECT customer#, SUBSTR(zip, 2, 3), INSTR(customer#, 1)
FROM customers;
SELECT customer#, SUBSTR(zip, 3, 2), INSTR(customer#, 3)
FROM customers;
Transcribed Image Text:Book Sales database (Just Lee Book sales Company): Which of below SQL option resolves this problem? Produce a list of each customer number and the third and fourth digits of his or her zip code. The query should also display the position of the first occurrence of a 3 in the customer number, if it exists. SELECT customer#, SUBSTR(zip, 3), INSTR(customer#, 3) FROM customers; SELECT customer#, INSTR(zip, 3, 2), INSTR(customer#, 3) FROM customers; SELECT customer#, SUBSTR(zip, 2, 3), INSTR(customer#, 1) FROM customers; SELECT customer#, SUBSTR(zip, 3, 2), INSTR(customer#, 3) FROM customers;
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