Queries Assignment Help | Project | Homework Problem Solution

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Queries are command that users sent to database in order to get the result. It is also called as line of inquiry. It can get the complete columns to get specific requirement based rows or column from the tables.

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Question:-The task of this assignment is to write the database queries to retrieve the data from the database of employees table in dual. The queries need to include Sjoins like left outer join etc.
Solution:-
Queries:
A.    SELECT employee1.emp_name, employee1.emp_SSN, dependents.firstname, dependents.birthdate,dependents.relation FROM employee1 LEFT JOIN dependents ON employee1.emp_Id=dependents.emp_id  ORDER BY employee1.emp_name;
B.    SELECT employee1.emp_name, proj_emp.proj_id FROM employee1 INNER JOIN proj_emp ON employee1.emp_id = proj_emp.emp_id  ORDER BY employee1.emp_name;
C.    Two Parts
a.     if you want to list all the employee whether have a mentor or not
select employee1.emp_name as employeeName, employee1.emp_SSN as employeeSSN, mentor.emp_name as mentorName, mentor.emp_SSN as mentorSSN from employee1  LEFT JOIN employee1 mentor  ON employee1.emp_sup = mentor.emp_id;
b.    If only want to list employees having mentors
select employee1.emp_name as employeeName, employee1.emp_SSN as employeeSSN, mentor.emp_name as mentorName, mentor.emp_SSN as mentorSSN from employee1 INNER JOIN employee1 mentor  ON employee1.emp_sup = mentor.emp_id;
D.    select temp1.empName, temp1.empSal, departments.dname FROM departments INNER JOIN(select temp.empName as empName, temp.empSal as empSal, temp.empDepId as empDep FROM (select employee1.emp_name as empName, employee1.emp_sal as empSal, employee1.emp_dep as empDepId FROM employee1 order by employee1.emp_sal desc) as temp GROUP BY temp.empDepId) as temp1 ON temp1.empDep = departments.dep_id;
E.    select sum(proj_emp.workhr), projects.proj_id from projects INNER JOIN proj_emp ON projects.proj_id = proj_emp.proj_id group by projects.proj_id order by sum(proj_emp.workhr) DESC LIMIT 1;
F.    select employee.emp_name, employee.emp_SSN, temp.NoOfDependents from employee INNER JOIN(select count(*) as NoOfDependents, dependents.emp_id as id from dependents group by dependents.emp_id order by count(*) DESC LIMIT 1) as temp ON temp.id = employee.emp_id;
G.    Two Parts:
a.    If you want to list employees who are doing projects
select temp.empName, temp.empDept, temp.projId, projects.pname from projects INNER JOIN (select employee1.emp_name as empName, employee1.emp_dep as empDept, proj_emp.proj_id as projId from employee1 LEFT JOIN proj_emp ON proj_emp.emp_id = employee1.emp_id) as temp ON temp.projId = projects.proj_id;
b.    If you want to list all the employees
select temp.empName, temp.empDept, temp.projId, projects.pname from projects RIGHT JOIN (select employee1.emp_name as empName, employee1.emp_dep as empDept, proj_emp.proj_id as projId from employee1 LEFT JOIN proj_emp ON proj_emp.emp_id = employee1.emp_id) as temp ON temp.projId = projects.proj_id;
H.    select departments.dname, temp2.finalTime from departments INNER JOIN(select sum(temp1.etime) as finalTime, temp1.empDept as depId FROM (select employee1.emp_dep as empDept, temp.empID as eID, temp.time as etime FROM employee1 INNER JOIN(select proj_emp.emp_id as empId, sum(proj_emp.workhr) as time from proj_emp group by proj_emp.emp_id) as temp ON employee1.emp_id = temp.empID) as temp1 group by temp1.empDept) as temp2 ON departments.dep_id = temp2.depID;