I have these two tables and I need to find the department name with maximum number of employees.
The other solutions were not for Oracle, so I'm posting this question. Also, it would be really helpful if the query can be explained thoroughly as I'm finding it hard to visualise it.
EMPLOYEE
EMPNO EMPNAME MANAGER SALARY DEPT_NO
1 Puja 6 30000 2
2 Purabi 1 15000 3
3 Barun 6 23000 2
4 Sudha 1 20000 1
5 Amal 2 20000 1
6 Rakesh 3 30000 4
DEPARTMENT
Dept_No Dept_Name Location
1 Production LaneA
2 Marketing LaneB
3 Sales LaneC
4 HR LaneD
So far I could manage getting the highest number of employees. So I was thinking if somehow I can write another sub-query where I count the employees in the departments again and compare them to the max_num_emp that I calculated in the first query.
This is the query which retrieves the maximum number of employees. It does not return the dept_no.
select count(dept_no)
from employee
group by dept_no
order by count(dept_no) desc
fetch first row only;
Expected output
DEPT_NAME
Production
Marketing
I can also add the dept_no column in the query, then I will have to somehow find out how to get the max and that was somehow giving me errors because the query was violating some rules. I had actually tried doing max(above query).
So I thought of just getting the maximum employee count and then determine all departments which have those many employees and display their name.
You have a working query which you need to join to the table department:
select d.Dept_Name
from department d inner join (
select dept_no
from employee
group by dept_no
order by count(*) desc
fetch first row only
) t
on t.dept_no = d.dept_no
Edit
Try this (I cannot try it):
select d.dept_name
from department d inner join (
select x.dept_no from (
select dept_no, rank() over (order by count(*) desc) rn
from employee
group by dept_no
order by count(dept_no) desc
) x
where x.rn = 1
) t
on t.dept_no = d.dept_no
You may have used FETCH..FIRST syntax using WITH TIES instead of ONLY.
SELECT d.dept_name
FROM department d
JOIN employee e ON d.dept_no = e.dept_no
GROUP BY d.dept_name
ORDER BY COUNT(*)
DESC FETCH FIRST 1 ROW WITH TIES ;
Demo
If you are not looking for duplicates, then:
select d.dept_name, count(*)
from department d join
employee e
on d.dept_no = e.dept_no
group by d.dept_no, d.dept_name
order by count(dept_no) desc
fetch first row only;
Related
I have a table with managers and the employees they manage as well as fName, lName etc. I need to count the number of employees a manager manages and return the maximum as well as all the rest of the information in the row.
SELECT TOP 1 Manager, COUNT (Manager) AS Manager_Count
FROM EMPLOYEE
GROUP BY Manager
ORDER BY COUNT(*) DESC
This is what I have but only returns the manager and the count. I have tried adding nested SELECT as well as SELECT * but just keep getting errors. Thanks for any help.
If you are talking about wanting all information for a given manager and the count of employees that manager has, I would suspect the manager is an employee, but also the manager is also a column per employees they are responsible for, such as:
EmployeeId
Name
Manager
1
Mary
0
2
Joe
1
3
Jane
1
4
Bob
2
Where Mary is the top-level manager and has two direct employees, but Joe is also a manager with one person under him. So your base query is correct
select top 1
e.Manager,
count(*) NumberOfEmployees
from
Employee e
group by
e.Manager
order by
count(*) desc
So this gives you the manager's employee ID and the count. Now, from this aggregated result, join back to the employee table for the manager's direct details.
select
e2.*,
Mgr.NumberOfEmployees
from
( select top 1
e.Manager,
count(*) NumberOfEmployees
from
Employee e
group by
e.Manager
order by
count(*) desc ) Mgr
Join Employee e2
on Mgr.Manager = e2.EmployeeId
I wrote a query joining two tables and I got a below resultset:
SELECT emp.employee_id,
dept.department_name,
dept.department_id
FROM employee emp,
department dept
WHERE emp.department_id = dept.department_id;
Employee_ID Department Department_ID
Mark Sales D1
Mark Marketing D2
Justin Textiles D3
Kimberley (null) (null)
However, I need to display below output with one new field called 'Status'.Mark can work in both the departments and so the count is "2" and the status will be 'Y' ( displaying of any one record is okay) . Justin works in only one department and count is 1 and status should be 'N'. Kimberley does not work anywhere and count is 0 and status should be 'N'.
Expected output:
Employee_ID Department Department_ID Status
Mark Sales D1 Y
Justin Textiles D3 N
Kimberley (null) (null) N
Please help.
I understand that you want to display the first department per user, and add a flag that indicates whether the employee belongs to at least one other department.
You can use window functions:
select
employee_id,
department_name,
department_id
case when cnt <= 1 then 'N' else 'Y' end status
from (
select
emp.employee_id,
dept.department_name,
dept.department_id,
row_number() over(partition by emp.employee_id order by dept.department_id) rn,
count(*) over(partition by emp.employee_id) cnt
from
employee emp
left join department dept on emp.department_id = dept.department_id
) t
where rn = 1
Side note: always use explicit joins (with the on keyword) instead of old-school, implicit joins (with commas in the from clause), whose syntax is harder to read and maintain.
I think this can be easily achieved using group by and keep clause with max as following:
SELECT emp.employee_id,
Max(dept.department_name) keep (dense_rank first order by dept.department_id) as department_name,
Max(dept.department_id) keep (dense_rank first order by dept.department_id) as department_id,
case when count(1) > 1 then 'Y' else 'N' end as status
FROM employee emp
LEFT JOIN department dept ON emp.department_id = dept.department_id
GROUP BY emp.employee_id;
Cheers!!
emp_name |salary
---------------
A |12568
B |3000
C |7852
D |2568
E |9852
F |1598
G |8569
I want a sql query to fetch the lowest 3 salaried employees
If you are using Oracle 12c or later you can make your query simpler with fetch. Instead of writing inner queries like this.
SELECT *
FROM
(SELECT * FROM EMPLOYEES EMP ORDER BY EMP.SALARY ASC
)
WHERE ROWNNUM <= 3
You can combine them into a single query.
SELECT * FROM employees emp ORDER BY emp.salary ASC
FETCH FIRST 3 ROWS ONLY;
More Information on the syntax and construct is available here.
http://www.dba-oracle.com/t_offset_fet_first_rows_only.htm
ORACLE:
SELECT emp_name
FROM ( SELECT *
FROM employees e
ORDER BY e.salary ASC)
WHERE ROWNUM < 4
Good luck!
You can use top 3 to get three record after ordering them in descending or ascending order. I have SQL server syntax but you can have idea from this for you target DBMS.
For top three max salaries
Select top 3 emp_name, salary
order by salary desc
For top three minimum salaries
Select top 3 emp_name, salary
order by salary asc
USE ASC AND LIMIT
Select emp_name, salary FROM TABLE_NAME
order by salary ASC LIMIT 3;
You didn't specify your DBMS, so this is ANSI SQL:
select emp_name, salary
from (
select emp_name, salary,
dense_rank() over (order by salary) as rnk
from employees
) t
where rnk <= 3;
This will also deal with employees that have the same salary. So the result might be more then three rows if more then one of the employees with the lowest salary have the same salary.
This question already has answers here:
How to find the employee with the second highest salary?
(5 answers)
Closed 2 years ago.
Well it is a well known question. Consider the below
EmployeeID EmployeeName Department Salary
----------- --------------- --------------- ---------
1 T Cook Finance 40000.00
2 D Michael Finance 25000.00
3 A Smith Finance 25000.00
4 D Adams Finance 15000.00
5 M Williams IT 80000.00
6 D Jones IT 40000.00
7 J Miller IT 50000.00
8 L Lewis IT 50000.00
9 A Anderson Back-Office 25000.00
10 S Martin Back-Office 15000.00
11 J Garcia Back-Office 15000.00
12 T Clerk Back-Office 10000.00
We need to find out the second highest salary
With Cte As
(
Select
level
,Department
,Max(Salary)
From plc2_employees
Where level = 2
Connect By Prior (Salary) > Salary)
Group By level,Department
)
Select
Employeeid
,EmployeeName
,Department
,Salary
From plc2_employees e1
Inner Join Cte e2 On e1.Department = e2.Department
Order By
e1.Department
, e1.Salary desc
,e1.EmployeeID
is somehow not working... I am not getting the correct result. Could anyone please help me out.
Something like
select * from
(
select EmployeeID, EmployeeName, Department, Salary,
rank () over (partition by Department order by Salary desc) r
from PLC2_Employees
)
where r = 2
Edit - tested it and it gives the answer you expected.
If you're going to teach yourself how to deal with CONNECT BY, you should first find a problem that is suited to the construct. CONNECT BY is meant for processing data that's in a hierarchical form, which your example is not. Salaries are not related to each other in a hierarchical fashion. Trying to force-fit a construct on the wrong problem is frustrating and doesn't really teach you anything.
Take a look at the classic employee-manager relationship in the demo HR schema you can install with Oracle. All employees report to a manager, including managers (except the top guy). You can then use this schema to create a query to show, for example, the Organization Chart for the company.
START WITH … CONNECT BY is designed to explore data that forms a graph, by exploring all possible descending paths. You specify the root nodes in the START WITH clause and the node connections in the CONNECT BY clause (not in the WHERE clause).
The WHERE clause filters will be processed after the hierachical conditions, same for GROUP BY and HAVING (of course because GROUP BY is computed after WHERE).
Therefore you MUST here CONNECT BY PRIOR department = department for example. You must also avoid that a node connection is done between two salaries when there is an intermediate salary.
Therefore the final query would resemble this:
SELECT level
, Department
, Salary
FROM plc2_employees pe1
START WITH pe1.salary = (select max(salary) from plc2_employees pe2 WHERE pe2.Department = pe1.Department)
CONNECT BY PRIOR pe1.Department = pe1.Department
AND PRIOR pe1.Salary > pe1.Salary
AND PRIOR pe1.Salary = ( SELECT MIN(Salary) FROM plc2_employees pe3
WHERE pe3.Department = pe1.Department
AND pe3.Salary > pe1.Salary
)
The recursion condition states that there is no intermediate salary between the child row and the parent row.
Note that this will really be unefficient…
Try this, it gives second highest salary
select MAX(Salary) as Salary
from Employee_salary
where Salary not in (select MAX(Salary) from Employee_salary)
You can use this query:
select * from
employee e1
where 2 = (select count (distinct (salary))
from employee e2
where e2.salary >=e1.salary);
find out second highest salary from employee table having column as salary:
Database : DB2
with t as
(
select distinct salary from employee order by salary desc
),
tr as
(
select salary, row_Number() over() r from t
)
select salary from tr where r = 2
Try this,
It gives second highest salary...
select MAX(Salary) as Salary
from Employee_salary
where Salary not in (select MAX(Salary) from Employee_salary )
If you want to find nth highest salary than you can use following query....
you need to do just one change.....
Put the value of N=nth highest
Cheers....:)
SELECT * FROM Employee_salary Emp1
WHERE (N-1) = (SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee_salary Emp2
WHERE Emp2.Salary > Emp1.Salary)
This will work -
SELECT MIN(Salary)
FROM employee
WHERE salary IN (SELECT TOP 2 salary FROM employee ORDER BY salary DESC)
First, select the distinct salaries in descending order (from greatest to least), from that set select the top 2 and put in ascending order (placing number 2 on top), then from those 2 select top 1:
select top 1 s.Salary
from
(select top 2 t.Salary
from
(select distinct Salary
from PLC2_Employees
order by Salary desc) t
order by Salary asc) s
How can I find the Nth highest salary in a table containing salaries in SQL Server?
You can use a Common Table Expression (CTE) to derive the answer.
Let's say you have the following salaries in the table Salaries:
EmployeeID Salary
--------------------
10101 50,000
90140 35,000
90151 72,000
18010 39,000
92389 80,000
We will use:
DECLARE #N int
SET #N = 3 -- Change the value here to pick a different salary rank
SELECT Salary
FROM (
SELECT row_number() OVER (ORDER BY Salary DESC) as SalaryRank, Salary
FROM Salaries
) as SalaryCTE
WHERE SalaryRank = #N
This will create a row number for each row after it has been sorted by the Salary in descending order, then retrieve the third row (which contains the third-highest record).
SQL Fiddle
For those of you who don't want a CTE (or are stuck in SQL 2000):
[Note: this performs noticably worse than the above example; running them side-by-side with an exceution plans shows a query cost of 36% for the CTE and 64% for the subquery]:
SELECT TOP 1 Salary
FROM
(
SELECT TOP N Salary
FROM Salaries
ORDER BY Salary DESC
) SalarySubquery
ORDER BY Salary ASC
where N is defined by you.
SalarySubquery is the alias I have given to the subquery, or the query that is in parentheses.
What the subquery does is it selects the top N salaries (we'll say 3 in this case), and orders them by the greatest salary.
If we want to see the third-highest salary, the subquery would return:
Salary
-----------
80,000
72,000
50,000
The outer query then selects the first salary from the subquery, except we're sorting it ascending this time, which sorts from smallest to largest, so 50,000 would be the first record sorted ascending.
As you can see, 50,000 is indeed the third-highest salary in the example.
You could use row_number to pick a specific row. For example, the 42nd highest salary:
select *
from (
select row_number() over (order by Salary desc) as rn
, *
from YourTable
) as Subquery
where rn = 42
Windowed functions like row_number can only appear in select or order by clauses. The workaround is placing the row_number in a subquery.
select MIN(salary) from (
select top 5 salary from employees order by salary desc) x
EmpID Name Salary
1 A 100
2 B 800
3 C 300
4 D 400
5 E 500
6 F 200
7 G 600
SELECT * FROM Employee E1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(E2.Salary))
FROM Employee E2
WHERE E2.Salary > E1.Salary
)
Suppose you want to find 5th highest salary, which means there are total 4 employees who have salary greater than 5th highest employee. So for each row from the outer query check the total number of salaries which are greater than current salary. Outer query will work for 100 first and check for number of salaries greater than 100. It will be 6, do not match (5-1) = 6 where clause of outerquery. Then for 800, and check for number of salaries greater than 800, 4=0 false then work for 300 and finally there are totally 4 records in the table which are greater than 300. Therefore 4=4 will meet the where clause and will return
3 C 300.
try it...
use table_name
select MAX(salary)
from emp_salary
WHERE marks NOT IN (select MAX(marks)
from student_marks )
Simple way WITHOUT using any special feature specific to Oracle, MySQL etc.
Suppose in EMPLOYEE table Salaries can be repeated.
Use query to find out rank of each ID.
select *
from (
select tout.sal, id, (select count(*) +1 from (select distinct(sal) distsal from
EMPLOYEE ) where distsal >tout.sal) as rank from EMPLOYEE tout
) result
order by rank
First we find out distinct salaries. Then we find out count of distinct salaries greater than each row. This is nothing but the rank of that id. For highest salary, this count will be zero. So '+1' is done to start rank from 1.
Now we can get IDs at Nth rank by adding where clause to above query.
select *
from (
select tout.sal, id, (select count(*) +1 from (select distinct(sal) distsal from
EMPLOYEE ) where distsal >tout.sal) as rank from EMPLOYEE tout
) result
where rank = N;
The easiest method is to get 2nd higest salary from table in SQL:
sql> select max(sal) from emp where sal not in (select max(sal) from emp);
Dont forget to use the distinct keyword:-
SELECT TOP 1 Salary
FROM
(
SELECT Distinct TOP N Salary
FROM Salaries
ORDER BY Salary DESC
) SalarySubquery
ORDER BY Salary ASC
Solution 1: This SQL to find the Nth highest salary should work in SQL Server, MySQL, DB2, Oracle, Teradata, and almost any other RDBMS: (note: low performance because of subquery)
SELECT * /*This is the outer query part */
FROM Employee Emp1
WHERE (N-1) = ( /* Subquery starts here */
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
The most important thing to understand in the query above is that the subquery is evaluated each and every time a row is processed by the outer query. In other words, the inner query can not be processed independently of the outer query since the inner query uses the Emp1 value as well.
In order to find the Nth highest salary, we just find the salary that has exactly N-1 salaries greater than itself.
Solution 2: Find the nth highest salary using the TOP keyword in SQL Server
SELECT TOP 1 Salary
FROM (
SELECT DISTINCT TOP N Salary
FROM Employee
ORDER BY Salary DESC
) AS Emp
ORDER BY Salary
Solution 3: Find the nth highest salary in SQL Server without using TOP
SELECT Salary FROM Employee
ORDER BY Salary DESC OFFSET N-1 ROW(S)
FETCH FIRST ROW ONLY
Note that I haven’t personally tested the SQL above, and I believe that it will only work in SQL Server 2012 and up.
SELECT * FROM
(select distinct postalcode from Customers order by postalcode DESC)
limit 4,1;
4 here means leave first 4 and show the next 1.
Try this it works for me.
Very simple one query to find nth highest salary
SELECT DISTINCT(Sal) FROM emp ORDER BY Salary DESC LIMIT n,1