The course I'm taking is database management. I have attached the necessary information for question. Please provide solution in detail so i can better understand why. I need solution query for part A and B utilizing SUBQUERY. Thank you in Advance! Part A - Find items with a unit price greater than the unit price of the item with item code ‘005’. Part B - Display all orders placed by customers from ‘curryst’. i.e., Customer Street should be 'curryst'.
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The course I'm taking is
I have attached the necessary information for question. Please provide solution in detail so i can better understand why. I need solution query for part A and B utilizing SUBQUERY. Thank you in Advance!
Part A - Find items with a unit price greater than the unit price of the item with item code ‘005’.
Part B - Display all orders placed by customers from ‘curryst’. i.e., Customer Street should be
'curryst'.
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- 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...Use Mysql code The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule for Feb 1, 2020 with the lesson date/time, student's first and last names, and the horse's registered name. Order the results in ascending order by lesson date/time, then by the horse's registered name. Make sure unassigned lesson times (student ID is NULL) appear in the results. Hint: Perform a join on the LessonSchedule, Student, and Horse tables, matching the student IDs and horse IDs.Write Mysql code for: The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule for Feb 1, 2020 with the lesson date/time, student's first and last names, and the horse's registered name. Order the results in ascending order by lesson date/time, then by the horse's registered name. Make sure unassigned lesson times (student ID is NULL) appear in the results. Hint: Perform a join on the LessonSchedule, Student, and Horse tables, matching the student IDs and horse IDs.
- Write Mysql code The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key RegisteredName Breed Height BirthDate Student with columns: ID - primary key FirstName LastName Street City State Zip Phone EmailAddress LessonSchedule with columns: HorseID - partial primary key, foreign key references Horse(ID) StudentID - foreign key references Student(ID) LessonDateTime - partial primary key Write a SELECT statement to create a lesson schedule with the lesson date/time, horse ID, and the student's first and last names. Order the results in ascending order by lesson date/time, then by horse ID. Unassigned lesson times (student ID is NULL) should not appear in the schedule. Hint: Perform a join on the Student and LessonSchedule tables, matching the student IDs.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 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,…
- DataBaseMid2021-2.doc [Compatibility Mode] - Word Sign in Table Tools File Home Insert Design Layout References Mailings Review View Help Design Layout O Tell me what you want to do 요 Share SECOND OUENIVAR The following Data Base shows some sample data values for an extended form of the Zoo and Visitors. The significance of Visiting row is that the Visitor visits a specified Zoo in a specified date. Zoo : Zno zi Visitors : City Gaza Nusirat Yno Name Address Age V1 Fuad Gaza 55 Z2 V2 Rami Gaza 20 Z3 Gaza V3 Ahmed Nusirat 31 V4 Ahmed Gaza 45 V5 Ali Rafa 20 Visiting: Vno Zno Z1 Date V2 11-11-1998 15-10-1999 V1 Z3 11-01-2003 11-01-2003 06-03-1997 V2 Z2 V2 Z3 V1 Z2 V3 Z1 06-031997 V3 Z3 11-11-2005 V4 Z3 11-11-2005 V2 Z2 12-12-2005 Given the following relational Algebra querya Visitors Þyisitora.Nno = Visiting.VnolIIxna.Znal Visiting) A IIzag [8 City = “Caza' ( Zoo)]1 A- Find the result relation of the above query B What was the request of the relational algebra query above (make a query…• Write a simple stored function that Takes a gender as an input ● Find the staff whose gender matches the input Return the results Write a simple stored procedure that Takes a gender as an input Find the staff whose gender matches the input • Print the resultsmysql> select * from order_db; | orderid | custid | itemcode | orderquantity | deliverydate 2 | 2 | 2 2 | 2 | 2 2 250 200 500 300 75 ~~~~N 2 2 B 2 5 rows in set (0.00 sec) mysql> select * from item_db; + | item_code | item_name | current_stock 10 33 | C 85 5 200 | 1 500 | 9 6 rows in set (0.00 sec) 800 600 125 200 265 100 | 300 250 | 550 | 300 | 300 2020-05-30 2021-09-21 | 2023-06-02 2020-05-29 2020-05-29 | unitprice | + 40 | 5| 150 650 15 | 100 | | | с | R | D | C | c modeofpayment | date_of_order | 2020-05-28 1998-02-15 2024-09-03 1998-02-15 1998-02-15
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