Oracle 12c: SQL
3rd Edition
ISBN: 9781305251038
Author: Joan Casteel
Publisher: Cengage Learning
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Chapter 5, Problem 1HOA
Add a new row in the ORDERS table with the following data: Order# = 1021, Customer# = 1009, and Order date = July 20, 2009.
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Task 2: Insert into the LARGE_PROPERTY table the office number, address, bedrooms, floors, monthly rent, and owner number from the PROPERTY table whose square footage is greater than 1,500 square feet.
Chapter 5 Solutions
Oracle 12c: SQL
Ch. 5 - Which command should you use to copy data from one...Ch. 5 - Which command can you use to change the existing...Ch. 5 - Which of the following is a correct statement? a....Ch. 5 - Which of the following is a valid SQL...Ch. 5 - Which of the following statements deletes all rows...Ch. 5 - What is the maximum number of records that can be...Ch. 5 - Add a new row in the ORDERS table with the...Ch. 5 - Prob. 2HOACh. 5 - Save the changes permanently to the database.
Ch. 5 - Add a new row in the ORDERS table with the...
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- Add a new row in the ORDERS table with the following data: Order# = 1023 and Customer# = 1009. Describe the error raised and what caused the error.arrow_forwardAdd a new customer row by using the sequence created in Question 1. The only data currently available for the customer is as follows: last name = Shoulders, first name = Frank, and zip = 23567.arrow_forwardModify the Job_class column of the EMPLOYEES table so that it allows storing a maximum width of two characters.arrow_forward
- he Horse table has the following columns: ID - integer, primary key RegisteredName - variable-length string Breed - variable-length string Height - decimal number BirthDate - date Write a SELECT statement to select the registered name, height, and birth date for only horses that have a height between 15.0 and 16.0 (inclusive) or have a birth date on or after January 1, 2020.arrow_forwardCreate table customer1( Customer# number primary key, First_name varchar2(25, Last_name varchar2(25)) Which insert statement is valid? 1. Insert into customer1 values (null, john, ‘smith’); 2. Insert into customer1 values(first_name, last_name) values (‘john’, ‘smith’); 3. Insert into customer1 (first_name, last_name, customer#) values (1000, ‘john’,’smith’); 4. Insert into customer1 values (1000, ‘john’, ‘smith’);arrow_forwardThe Horse table has the following columns: ID - integer, auto increment, primary key RegisteredName - variable-length string Breed - variable-length string, must be one of the following: Egyptian Arab, Holsteiner, Quarter Horse, Paint, Saddlebred Height - decimal number, must be ≥ 10.0 and ≤ 20.0 BirthDate - date, must be ≥ Jan 1, 2015 Make the following updates: Change the height to 15.6 for horse with ID 2. Change the registered name to Lady Luck and birth date to May 1, 2015 for horse with ID 4. Change every horse breed to NULL for horses born on or after December 22, 2016.arrow_forward
- The Horse table has the following columns: ID - integer, primary key RegisteredName - variable-length string Breed - variable-length string Height - decimal number BirthDate - date Write a SELECT statement to select the registered name and height for only horses that have an above average height. Order the results by height (ascending).arrow_forwardThe Horse table has the following columns: ID - integer, primary key RegisteredName - variable-length string Breed - variable-length string Height - decimal number BirthDate - date Write a SELECT statement to select the registered name and height for only horses that have an above average height. Order the results by height (ascending). Hint: Use a subquery to find the average height.arrow_forwardPlease give me correct solution.arrow_forward
- P5arrow_forwardCREATE TABLE employees ( employee_number int NOT NULL, last_name char(50) NOT NULL, first_name char(50) NOT NULL, salary int, dept_id int, CONSTRAINT employees_pk PRIMARY KEY (employee_number) ) ; INSERT INTO employees (employee_number, last_name, first_name, salary, dept_id) VALUES (1001, Smith', 'John', 62000, 500); INSERT INTO employees (employee_number, last_name, first_name, salary, dept_id) VALUES (1002, 'Anderson', 'Jack', 57500, 500); INSERT INTO employees (employee_number, last_name, first_name, salary, dept_id) VALUES (1003, 'Everest', 'Brad', 71000, 501); INSERT INTO employees (employee_number, last_name, first_name, salary, dept_id) VALUES (1004, Horvath', 'Jack', 42000, 501); PROBLEMS: 1. Create a select statement that will show all employees that have more than 50000 as their salary. 2. Create a select statement that will show all employees belonging into department 501. 3. Create a select statement that has a first name of Jack arranged them alphabetically in ascending…arrow_forwardChoose the SELECT statement that returns the given results from the Auto table below. ID Make Model Type Year Price 1 Toyota Camry sedan 2016 9800 2 Ford Escape crossover 2015 15900 3 Honda Civic sedan 2016 10200 4 Volkswagen Golf compact 2014 8800 5 Toyota RAV4 crossover 2016 12800 6 Toyota 4Runner suv 2015 16900 7 Honda CR-V crossover 2016 17900 Result : Honda 2 Toyota 3 Select Make, Count(Make) FROM Auto Group By Make; Select Make, COUNT(Make) From Auto Group By Make Having Count(Make) > 1; Select Make, Count(Make) From Auto Group By Make Having COUNT >1;arrow_forward
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