Refer to the sample tables and data below.Tables:people (id(pk), firstname, lastname, zip(fk))zip_code (zip(pk), city, state)mysql> select * from people;+--|id| firstname | lastname | zip1 Marty| 2 | Jennifer3Lorraine |4Biff| 5 | George5 rows in set (0.01 sec)McFly| ParkerMcFly45001 New York | NY45010 | New Jersey | NJ33647 TampaFL33765 | Clearwater | FL4 rows in set (0.01 sec)O rowsmysql> select * from zip_code;zip| city| state |3 rowsTannenMcFly6 rows9 rows45001| 3364733647How many rows will be returned by the following command?SELECT firstname, city, stateFROM people LEFT JOIN zip_codeON people.zip = zip_code.state;| 33647| 45010 |foreign key references zip_code(zip)

SummaryA SQL LEFT JOIN returns all rows from the left table, even though no matches found in the…

zip_code (zip(pk), city, state) mysql> select * from people; | id | firstname | lastname | zip 1 1 | Marty | McFly | 45001 2 Jennifer | Parker 3 | Lorraine | McFly | | 4 | Biff | 5 | George Tannen | McFly 5 rows in set (0.01 sec) 33647 | | 33647 | | 33647 | | 45010 |

Np Ms Office 365/Excel 2016 I Ntermed

1st Edition

ISBN:9781337508841

Author:Carey

Publisher:Carey

Chapter5: Working With Excel Tables, Pivottables, And Pivotcharts

Section: Chapter Questions

Problem 10RA

See similar textbooks

Similar questions

DROP TABLE IF EXISTS Worker; CREATE TABLE Worker ( ); 0 INSERT ); WORKER_ID INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, FIRST NAME TEXT, LAST NAME TEXT, SALARY INTEGER (15), JOINING DATE DATETIME, DEPARTMENT CHAR (25) INTO Worker (WORKER_ID, FIRST_NAME, LAST_NAME, SALARY, JOINING_DATE, DEPARTMENT) VALUES 'Arora', 100000, '14-02-20 09.00.00', 'HR'), (001, 'Monika', (002, 'Niharika', 'Verma' 80000, '14-06-11 09.00.00', 'Admin'), 'HR'), 'Admin'), Admin'), 'Account'), 'Kumar', 75000, '14-01-20 09.00.00', 'Account') DROP TABLE IF EXISTS Bonus; CREATE TABLE Bonus ( ); (003, 'Vishal' 'Singhal', 300000, '14-02-20 09.00.00', (004, 'Amitabh 'Singh', 500000, '14-02-20 09.00.00', (005, 'Vivek' 'Bhati', 500000, '14-06-11 09.00.00'. (006, 'Vipul', 'Diwan' 200000, '14-06-11 09.00.00', (007, 'Satish' (008, 'Geetika', 'Chauhan', 90000, '14-04-11 09.00.00', 'Admin'); WORKER_REF_ID INTEGER, BONUS AMOUNT INTEGER(10), BONUS_DATE DATETIME, FOREIGN KEY (WORKER_REF_ID) REFERENCES Worker (WORKER_ID) ON DELETE…
SQL Table: CREATE TABLE employees ( employee_id INT PRIMARY KEY, first_name VARCHAR(255), last_name VARCHAR(255), email VARCHAR(255) UNIQUE, phone_number VARCHAR(255), hire_date DATE, job_id VARCHAR(255), salary DECIMAL(10,2), commission_pct DECIMAL(5,2), manager_id INT, department_id INT, FOREIGN KEY (manager_id) REFERENCES employees(employee_id), FOREIGN KEY (department_id) REFERENCES departments(department_id)); CREATE TABLE departments ( department_id INT PRIMARY KEY, department_name VARCHAR(255), manager_id INT, FOREIGN KEY (manager_id) REFERENCES employees(employee_id)); INSERT INTO employees (employee_id, first_name, last_name, email, phone_number, hire_date, job_id, salary, commission_pct, manager_id, department_id)VALUES (1, 'John', 'Doe', 'john.doe@example.com', '555-555-5555', '2022-01-01', 'IT_PROG', 8000, 0.2, 3, 10), (2, 'Jane', 'Smith', 'jane.smith@example.com', '555-555-5556', '2022-02-01', 'IT_PROG', 9000,…
CREATE TABLE Users( C_Id varchar(8) NOT NULL, LastName Varchar(255) NOT NULL, FirstName varchar(255) NOT NULL, ema il varchar(255), Phone int, PRIMARY KÉY (C_Id) CREATE TABLE Restaurants( R_Id varchar(255) NOT NULL, Rname varchar(255) NOT NULL, Type varchar(255), Raddress varchar(255), Rphone int, Remail varchar(255), PRIMARY KEY(R_Id) ); CREATE TABLE RDate( T_num int NOT NULL, Bdate DATE, Rdate DATE, T_start varchar(255), E_start varchar(255), PRIMARY KEY(T_num) ); CREATE TABLE Book ingInfo( C_Id varchar(8) NOT NULL, R_Id int NOT NÚLL, PRIMARY KEY(C_Id, R_Id), FOREIGN KEY (C_Ià) REFERENCES Users(C Id), FOREIGN KEY (R_Id) REFERENCES Restaurant(R_Id) ); CREATE TABLE Rating( R_Id varchar(255) NÓT NULL, Rating Id varchar(255) NOT NULL, Rate varchar(255) CHECK (Rate>11), Comment LONG, PRIMARY KEY(Rating Id), FOREIGN KEY (R_Id) REFERENCES Restaurants(R_Id) ); Create 1 procedure Create 1 function Create 1 database trigger |
Update rows in Horse table The 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.
MySQL CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...6. Find the last name of all employees that were not hired on Tuesday *hint: (UseDATE_FORMAT() function amd the format you need is %W, also use UPPER()7. Find the number of employees in each department who have a manager.8. Find the number of employees for each manager whose employees' minimum salary is greaterthan 5000.9. Find the number of…
CREATE TABLE people (id INT NOT NULL AUTO INCREMENT PRIMARY KEY, firstname VARCHAR(30), lastname VARCHAR(30), spouse_id INT, CONSTRAINT spouse fk FOREIGN KEY (spouse id) REFERENCES people (id)); This is some sample data: +-- | id | firstname | lastname | spouse_id | 1 | Marty 2 | Jennifer 3 | Lorraine 4 | Biff 5 | George | McFly | Parker | McFly | Tannen | McFly NULL | NULL | 5 | NULL | | 3 |
CREATE TABLE DONORS (donor_id INT NOT NULL,donor_name varchar(255) NOT NULL,donor_age int,PRIMARY KEY (donor_id ));CREATE TABLE DONATIONS(pledge_id INT NOT NULL,pledge_date varchar(255) NOT NULL,amount_pledged int,Is_paid int,donor_id int,PRIMARY KEY (pledge_id ),FOREIGN KEY (donor_id) REFERENCES DONORS(donor_id)); 1. Populate every relation with three rows.
Update table product to set the value of column productType equal to the corresponding id value in the table productType. This requires a SELECT * FROM productType: to see what the id value is for each type. For example, in my database a. 1|meat b. 2| seafood c. 3| vegetable d. 4 grain and rice e. 5 bakery
MySql Workbench CREATE TABLE students ( id INT PRIMARY KEY, first_name VARCHAR(50), last_name VARCHAR(50), age INT, major VARCHAR(50), faculty VARCHAR(50)); CREATE TABLE location ( id INT PRIMARY KEY, name VARCHAR(50), rooms INT); CREATE TABLE faculty ( id INT PRIMARY KEY, name VARCHAR(50), department_id INT); 1. List last name of all students whose first name is longer than 4 letters in ascending order accordingto the last name. Duplicated rows should be removed from the output.2. Count the total number of rooms in Location.3. Find the number of students in each major.4. Find the number of employees in each department who get no commission or have salary less than5000.5. Find the maximum salary of employees in each department that the employee was hired 15 yearsbefore now. *hint: user TIMESTAMPDIFF(<unit type>,<Date_value 1>,<Date_value 2>), the unitcan be YEAR, MONTH, DAY, HOUR, etc...
WEEK SQL ASSIGNMENT Complete the following exercise: Alter your student table adding a new column called enroll_date using the datetime data type. Populate the new column (enroll_date) with the current date and time. Using your student table return the student first and last names concatenated with a space and then the enroll_date minus one month. Here is an example of the output: Name Date Jane Smith 2019-05-03 13:14:12 Tom Jones 2020-08-07 05:23:56 Using your student table return the month name for the enroll_date column. For example, if the enroll_date was 11-6-2019 then the month name would be June.
Create a new table containing these four columns: Emp#, Lastname, Firstname, and Job_class. The table name should be EMPLOYEES. The Job_class column should be able to store character strings up to a maximum length of four, but the column values shouldn’t be padded if the value has less than four characters. The Emp# column contains a numeric ID and should allow a five-digit number. Use column sizes you consider suitable for the Firstname and Lastname columns.
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