In the query below, what is the difference between INTEGER ad INTEGER(15)? Why is it used?CREATE TABLE Worker ( WORKER_ID INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, FIRST_NAME TEXT, LAST_NAME TEXT, SALARY INTEGER(15), JOINING_DATE DATETIME, DEPARTMENT CHAR(25));View keyboard shortcuts DROP TABLE IF EXISTS Worker;CREATE TABLE Worker ();0INSERT);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 CASCADEINSERT INTO Bonus(WORKER_REF_ID, BONUS_AMOUNT, BONUS_DATE) VALUES(001, 5000, '16-02-20'),(002, 3000,'16-06-11'),(003, 4000, '16-02-20'),(001, 4500, '16-02-20'),(002, 3500, '16-06-11');CREATE TABLE Title (WORKER_REF_ID INTEGER,WORKER_TITLE VARCHAR(64),AFFECTED FROM DATETIME,FOREIGN KEY (WORKER_REF_ID)REFERENCES Worker (WORKER_ID)ON DELETE CASCADEINSERT INTO Title(WORKER_REF_ID, WORKER_TITLE, AFFECTED_FROM) VALUES(001, 'Manager', '2016-02-20 00:00:00'),(002, 'Executive', '2016-06-11 00:00:00'),(008, 'Executive' '2016-06-11 00:00:00'),(005, 'Manager', '2016-06-11 00:00:00'),(004,'Asst. Manager', '2016-06-11 00:00:00'),(007, 'Executive', '2016-06-11 00:00:00'),(006,'Lead', '2016-06-11 00:00:00'),(003, 'Lead', '2016-06-11 00:00:00');I

INTEGER in SQL means the data to be stored in a particular field is an integer value. INTEGER…

n the query below, what is the difference between INTEGER ad INTEGER(15)? Why is it used? CREATE TABLE Worker ( WORKER_ID INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, FIRST_NAME TEXT, LAST_NAME TEXT, SALARY INTEGER(15), JOINING_DATE DATETIME, DEPARTMENT CHAR(25) ); View keyboard shortcuts

n the query below, what is the difference between INTEGER ad INTEGER(15)? Why is it used? CREATE TABLE Worker ( WORKER_ID INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, FIRST_NAME TEXT, LAST_NAME TEXT, SALARY INTEGER(15), JOINING_DATE DATETIME, DEPARTMENT CHAR(25) ); View keyboard shortcuts

Database Systems: Design, Implementation, & Management

12th Edition

ISBN:9781305627482

Author:Carlos Coronel, Steven Morris

Publisher:Carlos Coronel, Steven Morris

Chapter8: Advanced Sql

Section: Chapter Questions

Problem 3P

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herpever applicable 2. Title field in Instructor table should allow only Mr., Ms., Mrs., Dr., Prof. as values apply check constraints
CREATE TABLE sales ( SalesNumber INT(10) AUTO_INCREMENT PRIMARY KEY, SalesDate DATE, SalesTotal DECIMAL(10,2)); CREATE TABLE products ( number INT(11) AUTO_INCREMENT PRIMARY KEY, prodid VARCHAR(20) UNIQUE, prodname VARCHAR(30), price DECIMAL(10,2), onhand INT(11), CONSTRAINT fk_salesdetails_products FOREIGN KEY (prodid) REFERENCES salesdetails(prodid) ON DELETE RESTRICT); CREATE TABLE salesdetails ( number INT(10) AUTO_INCREMENT PRIMARY KEY, SalesNumber INT(10), prodid VARCHAR(20), price DECIMAL(7,2), qty INT(10), CONSTRAINT fk_salesdetails_sales FOREIGN KEY (SalesNumber) REFERENCES sales(SalesNumber) ON DELETE CASCADE, CONSTRAINT fk_salesdetails_products FOREIGN KEY (prodid) REFERENCES products(prodid) ON DELETE CASCADE); Using the above SQL and the schema diagram, create an ER diagram detailing all the fields of the tables and the relationship amount the tables.
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…
CREATE TABLE MEMBERSHIP ( MEM_NUM CHAR(3) CONSTRAINT MEMBER_MEM_NUM_PK PRIMARY KEY, MEM_FNAME VARCHAR(30) NOT NULL, MEM_LNAME VARCHAR(30) NOT NULL, MEM_STREET VARCHAR(30), MEM_CITY VARCHAR(10), MEM_STATE CHAR(2), MEM_ZIP CHAR(5), MEM_BALANCE NUMBER (2)); CREATE TABLE RENTAL ( RENT_NUM CHAR(4) CONSTRAINT RENTAL_RENT_NUM_PK PRIMARY KEY, RENT_DATE DATE, MEM_NUM CHAR(3), CONSTRAINT RENTAL_MEM_NUM_FK FOREIGN KEY (MEM_NUM) REFERENCES MEMBERSHIP); CREATE TABLE PRICE (PRICE_CODE CHAR(1) CONSTRAINT PRICE_PRICE_CODE_PK PRIMARY KEY, PRICE_DESC VARCHAR(20), PRICE_RENTFEE NUMBER (3,1), PRICE_DAILYATFEE NUMBER(3,1)); CREATE TABLE MOVIE (MOVIE_NUM CHAR(4) CONSTRAINT MOVIE_MOVIE_NUM_PK PRIMARY KEY, MOVIE_NAME VARCHAR(30) NOT NULL, MOVIE_YEAR CHAR(4), MOVIE_COST NUMBER(5,2), MOVIE_GENRE VARCHAR(15), PRICE_CODE CHAR(1), CONSTRAINT MOVIE_PRICE_CODE_FK FOREIGN KEY (PRICE_CODE) REFERENCES PRICE); CREATE TABLE VIDEO (VID_NUM CHAR(5) CONSTRAINT VIDEO_VIDEO_NUM_PK PRIMARY KEY, VID_INDATE…
1) Write the SQL code that will create the table structure for a table named EMP. This table is a subset of the EMPLOYEE table. The basic EMP table structure is summarized in the table below. (Note that the EMP_NUM is the PK)? ATTRIBUTE(FIELD) NAME DATADECLARATION Constraints EMP_NUM INT Auto generated Identifier, start from 101 EMP_LNAME VARCHAR(15) Not null EMP_FNAME VARCHAR(15) Not Null EMP_INITIAL CHAR(1) EMP_HIREDATE DATE Default value is system date JOB_CODE CHAR(3) Make sure input is in the range500,501,502
Refer to the film and inventory tables of the Sakila database. The tables in this lab have the same columns and data types but fewer rows. Write a query that lists the titles of films with the fewest rows in the inventory table. This query requires a subquery that computes the minimum of counts by film_id: SELECT MIN(count_film_id) FROM ( SELECT COUNT(film_id) AS count_film_id FROM inventory GROUP BY film_id ) AS temp_table; This subquery is provided in the template.
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.
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…
SQL: Start with the Employee table used in Lab 4. CREATE TABLE Employee ( EMPLOYEE_ID INT NOT NULL, FIRST_NAME CHAR(20), LAST_NAME CHAR(20), JOB_LEVEL INT, SALARY BIGINT, COMMISSION_PCT INT, CONSTRAINT PKEmployeeID PRIMARY KEY (EMPLOYEE_ID) ); INSERT INTO Employee VALUES (100,'Steven','King',1,50000,10), (101,'Neena','Kochhar',1,50000,10), (102,'Lex','De Haan',2,60000 ,15), (103,'Alexander','Hunold',2,60000,15), (104,'Bruce','Ernst',3,70000,25), (105,'Johann','Ernst',4,75000,30); Create another table. CREATE TABLE NumEmployeesByLastName ( LastName CHAR(20), NumEmployees INT ); a. Write a query to initialize the NumEmployeesByLastName table by inserting one row for each last name already in the Employee table, and having the accurate count of the number of employees having that last name. Assume that there are no null values in the Employee table. After your query, if you execute the following query, you would…
In Structure Query Information Select all columns from the jobs table using an asterisk.2. Select only job title, min salary, and max salary columns from the jobs table.3. Select all columns from the d_partners table by writing out all of the columns inthe select statement.4. Using the d_partners table, display the id, and then concatenate the first_nameand last_name together with a space in between the two columns. Give theconcatenated column an alias of NAME. Be sure to use the optional ASkeyword.5. Using the job_grades table, show the grade level, and then subtract thelowest_sal from the highest_sal to show the amount of variance there is for eachgrade level. Give the calculated column an alias of Salary_Variance. Do notuse the optional AS keyword.6. Using the employees table, show an unduplicated list of department_id’s.7. Change the SELECT statement you built for question 4, adding a single columncalled Partner_Information that concatenates the partner’s…
Add two columns to the EMPLOYEES table. One column, named EmpDate, contains thedate of employment for each employee, and its default value should be the system date. Thesecond column, named EndDate, contains employees’ date of termination. When I put this formal it says, line 2 has invalid identifier! Can you help me what I need to fix from line 2? alter table EMPLOYEES2add column EmpDate date = getdate(),add column EndDate;

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