Create a table named 'EMPLOYEE' with the required fields (empno int, first_name varchar(16), last_name varchar(16), hire_date date, salary int) and insert the following records given below. Sample Data: empno | first_name | last_name | hire_date | sal 1 | john | mathew | 2009-05-20 | 30000 2 | riya | roy | 2010-05-23 | 30000 || 2020-02-12 | 90000 || 3 | Leah | pv | Ensure that the 'EMPLOYEE' table contains, at the minimum, the provided records
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- CUSTOMER Customer table is composed of customer number, name and phone number. Give and fill-in the appropriate attribute name for each of the column. custID 123 124 125 126 reservelD 5001 5002 5003 5004 5005 5006 5007 RESERVATION Each reservation is for one taxi. Reservation table is composed of reservation identification number, start reservation date, end reservation date, reservation days requested by customers starting from reservation date until end of reservation date, customer number that make the reservation and taxi number assigned to the reservation. Give and fill-in the appropriate attribute name for each of the column. taxill custName Ahmad Bin Abdullah Fatimah Binti Adam LAI LA2 Ruqayya Binti Idris Sulaiman Bin Daud LA3 LA4 startDate 01/10/2019 05/10/2019 05/10/2019 15/10/2019 20/10/2019 27/10/2019 02/11/2019 taxiType endDate 03/10/2019 12/10/2019 08/10/2019 17/10/2019 25/10/2019 Sedan Sedan Van Van 30/10/2019 04/11/2019 cust Phoneno TAXI Taxi table is composed of taxi…Activity Description Given following two tables (EMPLOYEE, DEPARTMENT). Note, Dno in table EMPLOYEE and Dnumber in table DEPARTMENT all refer to the same thing (department number). CREATE TABLE EMPLOYEE ( Fname VARCHAR(15) Minit CHAR, Lname VARCHAR(15), Ssn CHAR(9), Bdate DATE, Address VARCHAR(30), Sex CHAR, Salary DECIMAL(10,2), Super_ssn CHAR(9), Dno INT, PRIMARY KEY (Ssn)); CREATE TABLE DEPARTMENT ( Dname VARCHAR(15), Dnumber INT, Mgr_ssn CHAR(9), Mgr_start_date DATE, PRIMARY KEY (Dnumber)); Write a single SQL statement to find the sum of the salaries of all employees of the ‘Engineering’ department, as well as the maximum salary, the minimum salary, and the average salary in this department. Based on Employee table, write a single SQL statement for each department, retrieve the department number, the number of employees in the department, and their average salary.Using the STUDENT table structure shown below, normalize the table to BCNF. Sample STUDENT Records Attribute Name Sample Value Sample Value Sample Value 199876 Sample Value 199877 STU_NUM 211343 200128 STU_LNAME Stephanos Smith Jones Ortiz STU_MAJOR Accounting Accounting Marketing Marketing DEPT_CODE ACCT ACCT MKTG MKTG DEPT_NAME Accounting Accounting Marketing Marketing DEPT_PHONE 4356 4356 4378 4378 COLLEGE_NAME Business Admin Business Admin Business Admin Business Admin ADVISOR_LNAME Grastrand Grastrand Gentry Tillery ADVISOR_OFFICE T201 T201 T228 T356 ADVISOR_BLDG Torre Building Torre Building Torre Building Torre Building ADVISOR_PHONE 2115 2115 2123 2159 STU_GPA 3.87 2.78 2.31 3.45 STU_HOURS 75 45 117 113 STU_CLASS Junior Sophomore Senior Senior Sample Value 223456 McKulski Statistics MATH Mathematics 3420 Arts & Sciences Chen J331 Jones Building 3209 3.58 87 Junior
- What is the proper way to insert this into my table? Insert into Vehicle values (0002,01234567890123452,'Chev','Trail-B', to_date(‘2011’,’yyyy’), 49000, 159000, to_date(’10-19-2020’,’mm-dd-yyy’),'10a'); CREATE TABLE Vehicle ( Vehicle_ID VARCHAR2(10), Vehicle_VIN VARCHAR2 (17) Unique, Vehicle_Make VARCHAR2 (20) NOT NULL, Vehicle_Model VARCHAR2 (20) NOT NULL, Vehicle_Year DATE NOT NULL, Vehicle_Purchase_Cost NUMBER (10,2), Vehicle_Mileage NUMBER (7), Vehicle_Mileage_Date DATE, Vehicle_Office VARCHAR2 (3), PRIMARY KEY (Vehicle_ID) );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…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
- 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,…Using the STUDENT table structure shown, do the following: Attribute Name Sample Value Sample Value Sample Value Sample Value Sample Value INV_NUM 211347 211347 211347 211348 211349 PROD_NUM AA-E3422QW QD-300932X RU-995748G AA-E3422QW GH-778345P SALE_DATE 15-Jan-2018 15-Jan-2018 15-Jan-2018 15-Jan-2018 16-Jan-2018 PROD_LABEL Rotary sander 0.25-in. drill bit Band saw Rotary sander Power drill VEND_CODE 211 211 309 211 157 VEND_NAME NeverFail, Inc. NeverFail, Inc. BeGood, Inc. NeverFail, Inc. ToughGo, Inc. QUANT_SOLD 1 8 1 2 1 PROD_PRICE $49.95 $3.45 $39.99 $49.95 $87.75 A. Write the relational schema, draw its dependency diagram and identify all dependencies, including all partial and transitive dependencies. You can assume that the table does not contain repeating groups and that any invoice number may reference more than one product. (Hint: This table…Consider Following Tables. Student ( regNo, sName, sCity, sCGPA, sAge) Course ( cCode, cName, credit_hours) Enrollment (regNo, cCode, semester, Grade) NOTE: you must use “JOINS” query only. Display names of all those students who have “A” Display student name and ccode of all those students who have ‘F’ grade in 5th Display data of all those students who have enrolled at least 10 Display grade of ‘Ali’ in course ‘CS101’. Display all courses of ‘Amir’ along with grade, also display names of courses which are not yet enrolled by ‘Amir’. Display course names along with number of enrolled student in it. If no one has yet enrolled that particular course, it must display ‘0’ in number of enrollment Display list of courses which are taught in 7th (course name must not repeat) Display list of all students by name and course name they have Display names of students and names of courses they have enrolled. Also display student’s name that have not enrolled any course Display all those students…
- 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…Q1- List all employees whose job_id is IT_PROG and hired after 01-JAN-95 and earning more than5000.Q2- List all employee except those with employee_id 101, 102, 103, 104 and 105Q3- Display the city attribute from locations table. Show the ones that contain “South” in the city.Sort the list in descending order.Q4- Display the first name, last name, salary, manager id and job id of all employees whose last fourcharacters of job id ends with “LERK”.Q5- Display the first name, last name and salary of all employees. Round the salary to the nearestunit of 1000 (ex. Salary=4500 New Salary=5000). Label the column as "New Salary".Add two columns to the EMPLOYEES table. One column, named EmpDate, contains the date of employment for each employee, and its default value should be the system date. The second column, named EndDate, contains employees’ date of termination