SELECT e.Last_Name, m.Last_Name AS "Manager Name", e.Salary*12 AS "Annual Salary", 1.City FROM employees e JOIN employees m ON e.Manager_id = m.Employee_Id JOIN departments d ON d.Department_id = e.Department_Id JOIN locations 1 ON 1.Location_id = d.Location_id WHERE 1.City "California"
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- in a shop ,there are 10 employee and 20 kinds of goods,goods id between 1-20 EMPLOYEE id first name last name gender 10001, 'Tom', 'Brown', 'F'10002, 'Elizabeth', 'Tremblay', 'F'10003, 'Gladys', 'Julie', 'F10004, 'John', 'Taylor', 'M10005, 'Amelia', 'Smith'10006, 'Logan', 'Katherine'10007, 'Leo', 'Brown'10007, 'Lem', 'Thompson'10009, 'Tom' 'Smith'10010, 'Emma', 'Campbell' ------------ and I want to add a name library in it ,like this how could i create a HTML file ,with will randomly create customers with these employee . there is a start button on the page . press "start "bottom ,and it will It will randomly match 10 items, customers, and goods, display goods id ,customername and gender ,employer name , id and gender . how to do such a page ?Publisher (name, phone, city), PK: name. Book (ISBN, title, year, published_by, previous_edition, price), PK: ISBN, FK: published_by refs Publisher, previous_edition refs Book. Author (SSN, first_name, last_name, address, income), PK: SSN. Write (aSSN, bISBN), PK: (aSSN, bISBN), FK: aSSN refs Author, bISBN refs Book. Editor (SSN, first_name, last_name, address, salary, works_for, book_count), PK: SSN, FK: works_for refs Publisher. Edit (eSSN, bISBN), PK: (eSSN, bISBN), FK: eSSN refs Editor, bISBN refs Book. Author_Editor (aeSSN, hours), PK: aeSSN, FK: aeSSN refs Author, aeSSN refs Editor. Give SQL statement for the following plain English language query based on the above schema. ---------------------------------------------------- Provide the title, year, and publisher name of every book and the first name and last name of the editor of the book. Note: publisher name is in Publisher table. so there are 4 tables that need to be traversed author, edit, editor, and then a publisher.…Customer ={idC(PK), surname} Contract ={idC(PK), customer, consultant, insurance_type, date} Consultant means: id of some Staff member Staff={idPerson(PK), surname,dob, boss, email} Dob means: date of birth, boss: is id of another Staff member supervisor Insurance_Type={idT(pk) , Tname , price , length } Length = number of months it is valid (from the date when it was bought ) Find consultants born after some contract was already signed?
- Publisher (name, phone, city), PK: name. Book (ISBN, title, year, published_by, previous_edition, price), PK: ISBN, FK: published_by refs Publisher, previous_edition refs Book. Author (SSN, first_name, last_name, address, income), PK: SSN. Write (aSSN, bISBN), PK: (aSSN, bISBN), FK: aSSN refs Author, bISBN refs Book. Editor (SSN, first_name, last_name, address, salary, works_for, book_count), PK: SSN, FK: works_for refs Publisher. Edit (eSSN, bISBN), PK: (eSSN, bISBN), FK: eSSN refs Editor, bISBN refs Book. Author_Editor (aeSSN, hours), PK: aeSSN, FK: aeSSN refs Author, aeSSN refs Editor. Give SQL statement for the following plain English language query based on the above schema. ---------------------------------------------------- Provide the title, year, and publisher name of every book and the first name and last name of the editor of the book. Note: publisher name is in Publisher table. so there are 4 tables that need to be traversed author,edit, editor, and then publisher.A Bank has many customers. Attribute for customer includes customer number, name, address (street, city, state and zip code), sex and date of birth. Customer can have multiple accounts. The bank does not allow join account (one account per customer). Type of account offer by the bank are Saving, Current and Fixed Deposit. Attribute for account are account no, account type, and account balance. Each of the account can only exist in a particular branch. Attribute for branch are branch no, branch name and address (street, city, state and zip code). Customer can open different account at different branch. Draw an ER diagram for the above situation. Identify the entity, attributes, primary key, relationship and multiplicity constraint in the ER diagram.Define the term " relational less than " .
- Publisher (name, phone, city), PK: name. Book (ISBN, title, year, published_by, previous_edition, price), PK: ISBN, FK: published_by refs Publisher, previous_edition refs Book. Author (SSN, first_name, last_name, address, income), PK: SSN. Write (aSSN, bISBN), PK: (aSSN, bISBN), FK: aSSN refs Author, bISBN refs Book. Editor (SSN, first_name, last_name, address, salary, works_for, book_count), PK: SSN, FK: works_for refs Publisher. Edit (eSSN, bISBN), PK: (eSSN, bISBN), FK: eSSN refs Editor, bISBN refs Book. Author_Editor (aeSSN, hours), PK: aeSSN, FK: aeSSN refs Author, aeSSN refs Editor. Give SQL statement for the following plain English language query based on the above schema. ---------------------------------------------------- Provide the title, year, and publisher name of every book and the first name and last name of the editor of the book.EXPERIMENTAL PROCEDURE Patients visit the hospital and their visit history is maintained by the hospital staff. Different physicians may be available on different dates. They diagnose and treat the patients of all categories. Some of treatments are free while others are to be paid by the patients. Sample data of the case is shown in the following chart. Patient History Report PatientID Name Address Visit Date Physician Diagnosis Treatment P-100809 A City: X 12-02-2007 20-02-2007 29-02-2007 15-03-2007 Dr. Z Dr. F Dr. R Dr. L Chest Infection Cold Hepatitis-A Eyes Infection Free Free Paid Paid P-200145 N City: Y 10-01-2007 15-02-2007 25-03-2007 Dr. L Dr. K Dr. A Bone Fracture Cough Flu Paid Free Free Task 1 Draw a dependency diagram and transform the above data to first normal form by eliminating repeating groups such that each row in the relation is atomic. Be sure to create an appropriate name for the…Q: Front office of any hotel is responsible for all room reservations, room allocations and fınal settlement of bills. Any company or person can reserve room for their future stay. They have to indicate from what date to what date they need the room. They also have to indicate how many rooms are required. Sometimes the reservations could be cancelled or dates or number of rooms changed. For reservation, cancelled or modification of rooms, customer receives an acknowledgement from the hotel. Draw sequence diagram for reserve room scenario, also draw the class diagram.
- Material : Relational Algebra Course : Database Systems Create a Relational Algebra From Entity Relational Diagram (ERD) Below for:a) Displays books published in the last 5 years from 2020.b) Displays details of customers who have done shopping (using the Join operator)c) Displays email customers who have done shopping (using the Set operator)d) Displays the ISBN book that has never been purchased (using the Set operator)C sharp Table: Student (the headers are the field names in the Students table) StudentID Name Age Gender ProgramID 791 Stephanie Brown 19 Female BCS 236 Shannon Dawn 25 Female BA 618 Geoff Berg 24 Male ARET 256 Andrew Schilling 22 Male BSC 902 Gary Sang 23 Male DAAD Note: There is a StudentDataSet with a Student table, a StudentTableAdapter, a StudentBindingSource, and a StudentDataGridView control on the form.Note: There is an Average query, named Average, that returns the average age of the student from the Student table.Note: There is also Max query, named Highest, that returns the highest age of the student from the Student table.Write the code you would place in the AverageButton click event on your form to call the Average query and the Highest query and display in a DifferenceLabel, the difference between the highest age of a student and the average age.J SHORTAND NOTATION FOR RELATIONAL SQL TABLES Notation Example Meaning Underlined A or A, B The attribute(s) is (are) a primary key Superscript name of relation AR or AR, BR The attribute(s) is (are) a foreign key referencing relation R As an example, the schema R(A, B, C, D, ES) S(F, G, H) corresponds to the following SQL tables: CREATE TABLE R ( A <any SQL type>, B <any SQL type>, C <any SQL type>, D <any SQL type>, E <any SQL type>, PRIMARY KEY(A), FOREIGN KEY (E) REFERENCES S(F) ); CREATE TABLE S ( F <any SQL type>, G <any SQL type>, H <any SQL type>, PRIMARY KEY(F)) EXERCISE Consider the following relational schema, representing five relations describing shopping transactions and information about credit cards generating them [the used notation is explained above]. SHOPPINGTRANSACTION (TransId, Date, Amount, Currency, ExchangeRate, CardNbrCREDITCARD, StoreIdSTORE) CREDITCARD (CardNbr, CardTypeCARDTYPE, CardOwnerOWNER, ExpDate, Limit)…