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Berlin
Asked: December 30, 20212021-12-30T21:04:09+05:30 2021-12-30T21:04:09+05:30In: DBMS

There are three relations in the following schema for the company: 1. EMPLOYEE (National_ID, Employee_Name, Salary, Service_Unit), 2. EMPLOYER (National_ID, …

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There are three relations in the following schema for the company: 1. EMPLOYEE (National_ID, Employee_Name, Salary, Service_Unit), 2. EMPLOYER (National_ID, Employer_Name, Salary, Service_Unit), 3. UNIT (Service_Unit, Contact_Phone_Number). Primary key has been underlined. The data stored in the schema has been shown as follows:

 

 

There are three relations in the following schema for the company: 1. EMPLOYEE (National_ID, Employee_Name, Salary, Service_Unit), 2. EMPLOYER (National_ID, ...

 

 

Write SQL commands for the following instructions one-by-one (24%, 4 points each):

1.     Please add a new record of employee with the National ID: P129799311, Name: Nohara Hiroshi, Service unit: Kasukabe Defence Force, Salary: $367,000.

2.     We would like to raise the salary of Isaac Netero to $65,000.

3.     Please list all the information of employee with the salary over $75,000 or between $30,000 and $40,000.

4.     Please calculate the average salary and the total number of employees in the company.

5.     If the company decides to raise the salary of the employers by 5%, please list the employers’ name and the new salary.

6. Please list all the employees’ name and order with the salary.

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    1. I'M ADMIN
      I'M ADMIN
      2021-12-30T21:08:17+05:30Added an answer on December 30, 2021 at 9:08 pm

      1) insert into employee values(‘P129799311′,’Nohara Hiroshi’,367000,’Kasukabe Defence Force’);

      2) update employer set salary=65000 where employer_name=’Isaac Netero’;

      3) select * from employee where salary>75000 or salary between 30000 and 40000;

      4) select avg(salary),count(national_id) from employee;

      5) select employer_name,salary*0.05 from employer;

      6) select * from employee order by salary;

      CREATE TABLES BY YOUR OWN

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