CREATE 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 INTOo 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 order.

CREATE 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 INTOo 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 order.

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 2.4CP

See similar textbooks

Similar questions

Add a new row in the ORDERS table with the following data: Order# = 1022, Customer# = 2000, and Order date = August 6, 2009. Describe the error raised and what caused the error.
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.
Add 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.
Modify the Job_class column of the EMPLOYEES table so that it allows storing a maximum width of two characters.
CREATE TABLE MOVIE( mov_id int NOT NULL, mov_title varchar(255) NOT NULL, mov_releaseDate date, gen_id int NOT NULL, PRIMARY KEY (mov_id), FOREIGN KEY (gen_id) REFERENCES GENRE(gen_id) ); CREATE TABLE GENRE( gen_id int NOT NULL, PRIMARY KEY (gen_id), gen_name varchar(255) NOT NULL );EXTRACT YEAR. Using the tables above, create a query that returns all the rows from the movie table and displays the movie id, movie title, and movie’s release year (not the full date). Must provide screenshots of the resulting table.
create table patients ( patient_id VARCHAR(255),name VARCHAR(255),insurance DECIMAL(15,2),date_admitted DATE,date_checked_out DATE,PRIMARY KEY(patient_id));INSERT INTO patients (ID , name , insurance , date_admitted , date_checked_out) values (2 , 'Ali' , '22.2' , '2/1/2020' , '3/1/2020')SELECT * FROM table patients create table doctor (doctor_id VARCHAR(255),name VARCHAR(255),specialization VARCHAR(255),PRIMARY KEY (doctor_id) ); create table test (test_id VARCHAR(255),test_name VARCHAR(255),test_date DATE,test_time TIME,result VARCHAR(255),PRIMARY KEY(test_id)); create table doctor_patient( patient_id VARCHAR(255), doctor_id VARCHAR(255),PRIMARY KEY(patient_id),FOREIGN KEY(doctor_id) REFERENCES doctor(doctor_id) ); create table test_log (test_log_id varchar(255),patient_id VARCHAR(255),test_id VARCHAR(255),doctor_id VARCHAR(255),comments VARCHAR(255),PRIMARY KEY(test_log_id),FOREIGN KEY(test_id) REFERENCES test(test_id),FOREIGN KEY(patient_id) REFERENCES…
create table patients ( patient_id VARCHAR(255),name VARCHAR(255),insurance DECIMAL(15,2),date_admitted DATE,date_checked_out DATE,PRIMARY KEY(patient_id));INSERT INTO patients (ID , name , insurance , date_admitted , date_checked_out) values (2 , 'Ali' , '22.2' , '2/1/2020' , '3/1/2020')SELECT * FROM table patients create table doctor (doctor_id VARCHAR(255),name VARCHAR(255),specialization VARCHAR(255),PRIMARY KEY (doctor_id) ); create table test (test_id VARCHAR(255),test_name VARCHAR(255),test_date DATE,test_time TIME,result VARCHAR(255),PRIMARY KEY(test_id)); create table doctor_patient( patient_id VARCHAR(255), doctor_id VARCHAR(255),PRIMARY KEY(patient_id),FOREIGN KEY(doctor_id) REFERENCES doctor(doctor_id) ); create table test_log (test_log_id varchar(255),patient_id VARCHAR(255),test_id VARCHAR(255),doctor_id VARCHAR(255),comments VARCHAR(255),PRIMARY KEY(test_log_id),FOREIGN KEY(test_id) REFERENCES test(test_id),FOREIGN KEY(patient_id) REFERENCES…
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…
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.
CREATE TABLE employees8(employee_id NUMBER(10) PRIMARY KEY,first_name VARCHAR(25) UNIQUE,last_name VARCHAR(25) UNIQUE,job_id NUMBER(3),salary NUMBER(6) NOT NULL,department_name VARCHAR2(20));CREATE TABLE jobs8(job_id NUMBER(3) PRIMARY KEY,job_title VARCHAR2(25),min_salary NUMBER(4)check(min_salary >=2900),max_salary NUMBER(6)check(max_salary<=100000)); CREATE TABLE departments8(department_id NUMBER(2) PRIMARY KEY,department_name VARCHAR2(30),location_name VARCHAR2(15) DEFAULT 'Istanbul'); how can I do this question according to these 3 tables ? Create a complex view that joining with three tables.
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.
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 between 10.0 and 20.0 BirthDate - date, must be on or after Jan 1, 2015 Insert the following data into the Horse table: RegisteredName Breed Height BirthDate Babe Quarter Horse 15.3 2015-02-10 Independence Holsteiner 16.0 2017-03-13 Ellie Saddlebred 15.0 2016-12-22 NULL Egyptian Arab 14.9 2019-10-12 Notes: Your SQL code does not display any results in Develop mode. Use Submit mode to test your code. In another lab that creates the Horse table, RegisteredName is NOT NULL. In this l