Refer to the sample tables and data below.Tables:people (id(pk), firstname, lastname, zip(fk)) -- foreign key references zip code(zip)zip_code (zip(pk), city, state)mysql> select * from people;| id | firstname | lastname | zip| 1 | Marty| McFly | 45001I 2Jennifer | Parker3 Lorraine | McFly+---3 rows in set (0.01 sec)O33647| 33647mysql> select from zip_code;| zip | city| state || NY45001 New York33647 TampaFL33765 Clearwater | FL3 rows in set (0.01 sec)What is returned by the following command?SELECT firstname, city, stateFROM people RIGHT 30IN zip codeON people.zip zip_code.zip;| firstname | city| Marty| Jennifer | Tampa| New York| Lorraine | Tampa| NULL+----4 rows in set (0.00 sec)FL| Clearwater | FL➖➖➖➖➖➖➖➖| firstname | cityNew YorkTampaMartyJennifer| Lorraine | Tampa| state || NY| FL+---3 rows in set (0.00 sec)| state || NYFL| FL Refer to the sample tables and data below.Tables:people (id(pk), firstname, lastname, zip(fk)) -- foreign key references zip code(zip)zip_code (zip(pk), city, state)mysql> select * from people;| id | firstname | lastname | zip| 1 | Marty| McFly | 45001I 2Jennifer | Parker3 Lorraine | McFly+---3 rows in set (0.01 sec)O33647| 33647mysql> select from zip_code;| zip | city| state || NY45001 New York33647 TampaFL33765 Clearwater | FL3 rows in set (0.01 sec)What is returned by the following command?SELECT firstname, city, stateFROM people RIGHT 30IN zip codeON people.zip zip_code.zip;| firstname | city| Marty| Jennifer | Tampa| New York| Lorraine | Tampa| NULL+----4 rows in set (0.00 sec)FL| Clearwater | FL➖➖➖➖➖➖➖➖| firstname | cityNew YorkTampaMartyJennifer| Lorraine | Tampa| state || NY| FL+---3 rows in set (0.00 sec)| state || NYFL| FL

Database which simply refers to the one that it is the database which contains the information that…

Refer to the sample tables and data below. Tables: people (id(pk), firstname, lastname, zip(fk)) foreign key references zip_code(zip) zip_code (zip(pk), city, state) mysql> select from people; | id | firstname | lastname | zip | 1 | Marty | McFly | 45001 | 2 | Jennifer | Parker 33647 | 3 | Lorraine | McFly | 33647 3 rows in set (0.01 sec) mysql> select from zip code; | zip | city | state | | 45001 | New York | NY FL | 33647 | Tampa | 33765 Clearwater | FL +-- 3 rows in set (0.01 sec) What is returned by the following command? SELECT firstname, city, state FROM people RIGHT 30IN zip code ON people.zip zip code.zip; firstname city | Marty Jennifer | Lorraine NULL | New York Tampa | Tampa | Clearwater +-- 4 rows in set (0.00 sec) | firstname | city | Marty New York Jennifer | Tampa | Lorraine Tampa 3 rows in set (0.00 sec) | state | | NY | FL | FL | FL | state | | NY FL | FL

Oracle 12c: SQL

3rd Edition

ISBN:9781305251038

Author:Joan Casteel

Publisher:Joan Casteel

Chapter6: Additional Database Objects

Section: Chapter Questions

Problem 6HOA: A new table has been requested to support tracking automated emails sent to customers. Create the...

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A new table has been requested to support tracking automated emails sent to customers. Create the table and add data as described below. Tablename: email_log Columns: emailid (numeric), emaildate (datetime), customer# (numeric) Primary key: emailid column, define as an Identity Column Add the following data rows and display resulting rows (if any errors occur, explain why the error is expected) Emaildate = current date, customer# = 1007 Emailid = specify to use the column default value, emaildate = current date, customer# = 1008 Emailid = 25, emaildate = current date, customer# = 1009
Examine this product table's column definitions:pid number primary keypname varchar2(50)You must display column headings as shown:Product_ID Product NameWhich SELECT statement does this? SELECT pid Product_ID, pname Product Name FROM product; SELECT pid Product_ID, pname "Product Name" FROM product; SELECT pid @Product_ID, pname @"Product Name" FROM product; SELECT pid AS Product_ID, pname AS Product Name FROM product
Database course: Which of the following queries is used to list a unique value for V_CODE, where the list will produce only a list of those values that are different from one another? SELECT UNIQUE V_CODE FROM PRODUCT; SELECT DISTINCT V_CODE FROM PRODUCT; SELECT DIFFERENT V_CODE FROM PRODUCT; SELECT ONLY V_CODE FROM PRODUCT;
35. Modify the MODEL table to add the attribute and insert the values shown in the following table. Ch08_AviaCo_MySQL
SQL Please see the attached image.
SQL script which is the physical design for this database ### Animals This table contains the following data about all animals in the zoo. - animal ID (primary key) - animal name - animal birthdate - species - habitat (foreign key) [1 habitat to many relationship] ### Habitats This table contains data about the habitats in the zoo. Each animal lives in a specific habitat. The habitat table should include the following data - habitat name, foreign key) - habitat location ### Database CREATE DATABASE zoo; Write queries to create the tables. Be sure to include primary keys for each table and the required foreign keys and referential integrity actions. Add appropriate indexes. This database requires at least 2 indexes in addition to those automatically created by MySQL.
The database should contain the following six tables: Artists (artist_id: char(30), artist_name: char(30), artist_pop: int(5))Tracks (track_id: char(30), track_name: char(30), duration: int(10), tempo:real) Record (artist_id: char(30), track_id: char(30))Users (user_id: char(30), user_name: char(30), age: int(5), nationality: char(30), num_track_listened: int(10))Listen (user_id: char(30), track_id: char(30))Follow (user_id: char(30), artist_id: char(30), follow_date: datetime) Write a single SQL query for each request below. Find the ids of those users who have listened to at least one track but have not followed any artists. Find the ids and names of those users who have listened to some tracks recorded by the artist named 'Adele'. Find the ids and names of those users who have not listened to any track recorded by the artist named 'Adele'. Use ''NOT EXISTS'' to answer this query.
The following fields are presented in the table: Member Last Name, MemberFirstName, Street, City, State, ZipCode and MemberFee. The table contains 75,000 documents. How would you build indexes for the table and why would you create these indexes?
Having this information. Students Attribute Name Data Type StudentID char(11) FirstName varchar(20) LastName varchar(20) Gender char(1) DateofBirth date Courses Attribute Name Data Type CourseCode varchar(6) CourseName varchar(70) Level char(2) Credits int Registration Attribute Name Data Type StudentID char(11) CourseCode varchar(6) Grade decimal(2, 1) Write SQL code to: 1. Find the youngest student’s name. 2. How many students have NOT taken the course ‘CS056’? 3. Change the credit for all Computer Science courses (i.e. course codes CS***) with five credits from 5 to 6. 4. Delete the registration record for the student 908409-0010.
1. Use the following table to answer the questions below: Table name: Products Column Data Type Length Constraint Role Name Stores the ID of a product P ID P_Name Primary key Number Stores the name of a Varchar2 25 product P_entryDate Stores the entry date of a product Date Cannot be null P_Expiry Stores the expiry date of a product Date P_Еxpiry date should be greater than P_entryDate P_Price Stores the price of a product Number 3,2 Cannot be null a. Write a SQL query that displays the mumber of products that will expire in the year 2021 b. Write in SQL a query that displays the ids of the most expensive products c. Create a view named ProductDetails' to hold all the details of products whose names start with 'DELL' and their prices are between 200 and 500 Omani Rials
Having this information. Students Attribute Name Data Type StudentID char(11) FirstName varchar(20) LastName varchar(20) Gender char(1) DateofBirth date Courses Attribute Name Data Type CourseCode varchar(6) CourseName varchar(70) Level char(2) Credits int Registration Attribute Name Data Type StudentID char(11) CourseCode varchar(6) Grade decimal(2, 1) Write SQL code to: 1. Find the students who have not registered any course yet. 2. Find the student (studentID only), the sum of grade, and the average of grade, for each student who has registered courses and the average of grade is below 3. 3. Find the student who has registered the most number of courses. If there are multiple students who have registered the most number of courses, just take one student (To take only one record - the 1st record in the returned result, use ‘limit 1’ at the end of your SQL code). 4.Find the students (studentID only) who have taken either “PHY210” or “BIO001”.
Given the following MySQL table: persons (The Primary key in the first column and it is an automatic number) Table name: wilayats (for Wilayat data) wsq Table name: persons column names are highlighted prid prName prSalary prDoB prGender prWilyatsq wName Person ID Gender Wilayat id Salary Birth date 630 2011-08-03 Muneer Khalid 980 2009-01-25 870 2019-03-14 Name Wilayat ID Wilayat Name Nasser Aziz Nizwa Izki Bahla 200 M 3 300 M 2 Rawan Sami 500 F 3 2. Given also the following HTML form page and its source code that is used to add data to MYSQL database Web Browser Output 1. 2. Name :br> 3. Salary:br> 4. Birth Date: 5. Wilayat: 6. -- Select -- 7. Bahla.. 8. Gender:Male Adding data for a new Person Name: Jamal Obaid Salary: 550 Birth Date: 12/20/2002 Wilayat Bahla Gender: O Male O Female 10. Female 11. Save Data Note the following: • Do NOT write the validation part. Assume that data was input correctly. • Assume that the database connection id is $conn Write PHP code to do the…