I have a employee hive table with columns name and department. where 1 employee can belongs to multiple departments.
name, department
xxx,finance
xxx,hr
xxx,transport
xxx,sale
yyy,finance
yyy,hr
yyy,transport
zzz,finace
zzz,hr
zzz,transport
zzz,sale
I need to know distinct employee name who does not belongs to "sale" department.
As of hive 0.13
Select name from employee
where employee.name not in
(select name from employee where department = 'sale')
group by name;
Hopefully names are unique across employees.
You could write a subquery to pull all names that are in sales. Then join that query's results back to your table.
select
results.name,
results.department
from
(select e.name
from employee e
where e.department='sale' group by e.name) invalid_names
right join
(select
e.name,
e.department
from employee e) results
on invalid_names.name = results.name
where invalid_names.name is null;
I'd imagine there is a better way to do this, but this should work :)
Related
I am learning postgresql and Inner join I have following table.
Employee
Id Name DepartmentId
1 John S. 1
2 Smith P. 1
3 Anil K. 2
Department
Department
Id Name
1 HR
2 Admin
I want to query to return the Department Name and numbers of employee in each department.
SELECT Department.name , COUNT(Employee.id) FROM Department INNER JOIN Employee ON Department.Id = Employee.DepartmentId Group BY Employee.department_id;
I dont know what I did wrong as I am new to database Query.
When involving all rows or major parts of the "many" table, it's typically faster to aggregate first and join later. Certainly the case here, since we are after counts for "each department", and there is no WHERE clause at all.
SELECT d.name, COALESCE(e.ct, 0) AS nr_employees
FROM department d
LEFT JOIN (
SELECT department_id AS id, count(*) AS ct
FROM employee
GROUP BY department_id
) e USING (id);
Also made it a LEFT [OUTER] JOIN, to keep departments without any employees in the result. And COALESCE to report 0 employees instead of NULL in that case.
Related, with more explanation:
Query with LEFT JOIN not returning rows for count of 0
Your original query would work too, after fixing the GROUP BY clause:
SELECT department.name, COUNT(employee.id)
FROM department
INNER JOIN employee ON department.id = employee.department_id
Group BY department.id; --!
That's assuming department.id is the PRIMARY KEY of the table, in which case it covers all columns of that table, including department.name. And you may want LEFT JOIN like above.
Aside: Consider legal, lower-case names exclusively in Postgres. See:
Are PostgreSQL column names case-sensitive?
imagine I have two tables, the "departments" table and the "employee" table.
This employee table has a column for "category".
I'd like to make a query for selecting departments that only have a specified type of employees.
Thank you.
You will need to perform a join from your departments and employee table on whatever columns link these two tables together. In the where clause, you will specify what types of employees that you want.
This will return a row for each employee, which might not be what you want. You may use the distinct function on the important columns that you're looking for in the departments table to get the final answer.
select distinct dept_id
from employee
where category = 'cat1'
and dept_id not in (select distinct dept_id
from employee
where dept_id <> 'cat1');
SELECT dept_id
FROM departments
WHERE dept_id NOT IN
(SELECT DISTINCT dept_id
FROM employee
WHERE category_id != #specified_category)
This query assumes there are no departments with no employees, since it will also return those empty departments. If that's a problem, you can add:
AND dept_id IN (SELECT distinct dept_id FROM employee)
Select d.id_department from departments d where not exists
(Select e.id_employee from employees e where e.category!=your_category and e.id_department=d.id_department) you also need to verify that department has employees.
I'm trying to figure out a query which shows the names of the employees who worked in more than 2 departments along with their wage and contact details. I have two tables employees and department. Both of these having the EmployeeName field. I know we have to use the Count function but don't really know how to create the query.
here the tablename and Fields:
Employee (employeeName, wage, contactNo)
Department (employeeName, departmentNo, hours, startDate)
You SQL query would be the following
SELECT e.employeeName, count(departmentNo) FROM Employee e
INNER JOIN Department d ON e.employeeName=d.employeeName
GROUP BY e.employeeName
HAVING COUNT(departmentNo)>2
you can use following query:
SELECT e.employeeName, count(d.departmentname)
FROM Employee e, Department d
where e.deptid=d.deptid
GROUP BY e.employeeName
HAVING COUNT(e.deptid)>=2
I am trying to write a query that lists the name of a manager and the number of people they manage.
In the Manager table we have the managers name and id.
In the Employee table we have the employees name, id and managerID.
I don't understand how to get the count of the employees that a manager manages.
SELECT COUNT(e.EmpID), m.ManagerID
FROM Employee e
INNER JOIN Manager m
ON e.ManagerID= m.ManagerID
GROUP BY m.ManagerID
SELECT m.Name, COUNT(e.id) AS NumberOfEmployeesManaged
FROM Manager m INNER JOIN Employee e ON m.id = e.managerID
GROUP BY m.Name
That should do it I think, just a simple count of the employee ids after joining the manager and employee tables, grouped on manager name.
SELECT count(emp.empid), mgr.managerid
FROM Employee emp
INNER JOIN Manager mgr ON emp.managerid=mgr.managerid
GROUP BY mgr.managerid;
I don't know if you can use the COUNT aggregator in a JOIN. But you can run 2 queries. One would select the manager's name & id. The 2nd would look like this:
$id = the manager's id
SELECT COUNT(*) FROM Employee WHERE managerID=$id
Alternately, you could not use COUNT and run a query like this:
SELECT id FROM Employee WHERE managerID=$id
Then the # of resulting rows would be the count of employees managed by the manager.
So I have a table that has, employee number, employee name, supervisor number.
I want to run a query that will retrieve employee name, employee number, supervisor name and supervisor number. Only one employee doesn't have a supervisor meaning it will have to display nulls. How would I do this? I'm using Oracle SQL Plus. My attempts haven't worked at all! Any help would be much appreciated.
SELECT ename Employee, empno Emp#, super Manager#
FROM emp;
That gets me three of the columns but to be honest I don't even know where to start to get the supervisors names.
It's for university, but I'm studying for a test it's not for an assignment so no cheating happening here :).
The following should work, and give you nulls if the employee has no supervisor:
SELECT empGrunt.ename Employee
, empGrunt.empno EmpNum
, empSuper.ename SupervisorName
, empSuper.empno SupervisorName
FROM emp empGrunt LEFT OUTER JOIN emp empSuper
ON empGrunt.super = empSuper.empno
Assuming that SupervisorNumber is a foreign key relationship back to the Employee table (where it's the EmployeeNumber of the supervisor's record), then you need to use an outer join.
What you need in this case is a left join:
select
e.EmployeeName,
e.EmployeeNumber,
s.EmployeeName as SupervisorName
from Employee e
left join Employee s on s.EmployeeNumber = e.SupervisorNumber