Find 2nd highest salary in every department - sql-server

Hi I have a schema Employee(Employeeid,Name,departmentid,salary). I want to find out 2nd highest salary in each department.
select DepartmentID,name,salary from
(select Departmentid,name,salary, rank() over
(partition by departmentid order by salary desc)as
Rank from employee)t where t.Rank=2;
This does the job but if there is only 1 employee in a department then it does not printout that salary. Can anyone please help me with that?

Try this
Use Count() over() analytic function to count the records in each department. When the count is 1 then take Rank 1
SELECT departmentid,
NAME,
salary
FROM (SELECT departmentid,
NAME,
salary,
Dense_rank()OVER (partition BY departmentid
ORDER BY salary DESC) AS Rank,
Count(1)OVER(partition BY departmentid) AS cnt
FROM employee)t
WHERE t.rank = 2
OR ( t.rank = 1
AND cnt = 1 )
Note : I have used DENSE_RANK over RANK because, when there is a TIE in first salary you will not get RANK = 2

Related

second highest salary in each department

I am trying to find the second highest salary in each department.
Schema:
CREATE TABLE employees
(
ID int NOT NULL,
NAME char(50) NOT NULL,
departmentid int,
salary int
);
Sample records:
/*departmentid =1 */
INSERT INTO employees VALUES (1, 'Max', 1, 90000);
INSERT INTO employees VALUES (2, 'Joe', 1, 70000);
INSERT INTO employees VALUES (3, 'Randy', 1, 70000);
/*departmentid =2 */
INSERT INTO employees VALUES (4, 'Henry', 2, 80000);
INSERT INTO employees VALUES (5, 'SAM', 2, 60000);
/*departmentid =3 */
INSERT INTO employees VALUES (6, 'Janet', 3, 69000);
My query:
SELECT departmentid,
NAME,
salary
FROM
(
SELECT
departmentid,
NAME,
salary,
Dense_rank()OVER (partition BY departmentid
ORDER BY salary DESC) AS Rank,
Count(1)OVER(partition BY departmentid) AS cnt
FROM
employees
)t
WHERE
t.rank = 2
OR ( t.rank = 1
AND cnt = 1 )
The output I am getting is as below;
departmentid NAME salary
1 Joe 70000
1 Randy 70000
2 SAM 60000
3 Janet 69000
My expected output is
departmentid NAME salary
1 Joe 70000
1 Randy 70000
2 SAM 60000
3 NULL NULL
As there is only one record for departmentid=3, it should return null.
What is wrong with this query? Any other ways to achieve this result?
I've also included a SQL fiddle.
ROW_NUMBER() and select = 2
;WITH salary AS
(
[RN] = SELECT ROW_NUMBER() OVER (PARTITION BY departmentid ORDER BY salary),*
FROM <table>
)
SELECT
*
FROM salary
WHERE [RN] = 2
I've used two CTEs.
The first returns a list of every department. You'll need this to ensure departments with less than 2 salaries are included in the final result.
The second ranks each employee within their department.
Finally, I've used a left outer join to maintain the complete list of departments.
WITH Department AS
(
-- Returns a list of the departments.
SELECT
departmentid
FROM
employees
GROUP BY
departmentid
),
EmployeeRanked AS
(
SELECT
DENSE_RANK() OVER (PARTITION BY departmentid ORDER BY salary DESC) AS [Rank],
departmentid,
NAME,
salary
FROM
employees
)
SELECT
er.Rank,
d.departmentid,
er.NAME,
er.salary
FROM
Department AS d
LEFT OUTER JOIN EmployeeRanked AS er ON er.departmentid = d.departmentid
AND er.[Rank] = 2
;
Returns
Rank departmentid NAME salary
2 1 Joe 70000
2 1 Randy 70000
2 2 SAM 60000
(null) 3 (null) (null)
Use a sub query as i wrote here : http://sqlfiddle.com/#!6/bb5e1/26
with ranks as(
SELECT departmentid,
salary,
row_number() over (partition by (departmentid) order by salary desc) as rank
FROM employees
)
Select *
from ranks
Where ranks.rank = 2
If the departmentid having only one row, and if you consider that also. Then
Query
;with cte as(
select [rank] = dense_rank() over(
partition by departmentid
order by departmentid, salary desc
), *
from employees
)
select ID, NAME, departmentid, salary from cte
where [rank] = 2
union all
select max(ID), max(NAME), departmentid, max(salary)
from cte
group by departmentid
having count([rank]) = 1;
There is also a simple way:
SELECT TOP 1 * FROM (Select top 2 * FROM employees order by salary desc ) e Order by salary asc
Edit: this returns only the 2nd highest overall
I think you can get correct answer by just removing below code from your code
OR ( t.rank = 1
AND cnt = 1 )
also main table should be left join from this result to get null in rest of columns

SQL Server query should return max value records

I have table like this:
id_Seq_No emp_name Current_Property_value
-----------------------------------------------
1 John 100
2 Peter 200
3 Pollard 50
4 John 500
I want the max record value of particular employee.
For example, John has 2 records seq_no 1, 4. I want 4th seq_no Current_Property_Value in single query.
Select
max(id_Seq_No)
from
t1
where
emp_name = 'John'
To get the Current_Property_value, just order the results by id_Seq_No and get the first one:
SELECT
TOP 1 Current_Property_value
FROM
table
WHERE
emp_name = 'John'
ORDER BY
id_Seq_No DESC
this will give highest for all tied employees
select top 1 with ties
id_Seq_No,emp_name,Current_Property_value
from
table
order by
row_number() over (partition by emp_name order by Current_Property_value desc)
You can use ROW_NUMBER with CTE.
Query
;WITH CTE AS(
SELECT rn = ROW_NUMBER() OVER(
PARTITION BY emp_name
ORDER BY id_Seq_No DESC
), *
FROM your_table_name
WHERE emp_name = 'John'
)
SELECT * FROM CTE
WHERE rn = 1;

How to select last id(where id >0) fod each group in SQL Server 2008

I want to select last record of price column of my table where my value of column is greater than zero. How can I do that?
My stored procedure is :
SELECT
id, name, price
FROM
messages
WHERE
id IN (SELECT MAX(id)
FROM messages
WHERE price > 0
GROUP BY name)
The problem is that this code select max id that price is greater than zero not last id. Means select id=2 and id=6
But in last id of group (frank) price is zero but this stored procedure select id=2 while I want stored procedure select only id =6
id name price
--------------
1 frank 1000
2 frank 500
3 frank 0
4 john 200
5 john 100
6 john 20
There are multiple ways to approach this. Following your method, though, you just need to move the price comparison to the outer query:
SELECT id, name, price
FROM messages
WHERE price > 0 AND
id IN (SELECT MAX(id)
FROM messages
GROUP BY name
);
I would be more inclined to write this as:
select m.*
from (select m.*,
row_number() over (partition by name order by id desc) as seqnum
from messages m
) m
where seqnum = 1 and price > 0;
with cte as (
select row_number() over(partition by id order by id desc) rn,
id,name,price
from messages
)
select id,name,price
from cte
where rn=1 and price>0
Something like this.

SQL - How can I Group sets of sequential numbers and return the MIN and Max Dates

this is driving me crazy! does anyone know how to write some SQL that will return the MIN and MAX dates from groups of sequential numbers? please see screen shots below.
This is the SQL I used:
SELECT
num
, empid
, orderdate
FROM
(SELECT
ROW_NUMBER() OVER (ORDER BY orderdate) AS Num
, empid
, orderdate
FROM TSQL.Sales.Orders)T1
WHERE empid = 4
This is what it returns:
What I would like to do is get the Min and Max dates for each set of sequential numbers based on the num column. For example: the first set would be num 3, 4, 5 & 6. so the Min date is 2006-07-08 and the Max date is 2006-07-10
See example of results needed below
Any help with this would be much appreciated, thank you in advance
Update
I have now changed the SQL to do what I needed: example as follows:
Select
empid
, Island
, MIN(orderdate) as 'From'
, Max(orderdate) as 'To'
From
(select
empid
, num
, num - ROW_NUMBER() OVER (ORDER BY num, orderdate) as Island
, orderdate
from
(Select
ROW_NUMBER() OVER (ORDER BY orderdate) as Num
, empid
, orderdate
from TSQL.Sales.Orders)T1
where empid = 4
)T2
group By
empid
, Island
Result
Thank you so much for your help on this, I have been trying this for ages
Regards
Jason
This should do it:
;with dateSequences(num, empId, orderDate) as
(
select ROW_NUMBER() over (order by orderdate) as num
, empId
, orderdate
from yourTable
),
dateGroups(groupNum, empId, orderDate, num) as
(
select currD.num, currD.empid, currD.orderDate, currD.num
from dateSequences currD
left join dateSequences prevD on prevD.num = currD.num - 1 and prevD.empid = currD.empId
where prevD.num is null
union all
select dg.groupNum, d.empId, d.orderDate, d.num
from dateSequences d
inner join dateGroups dg on dg.num + 1 = d.num and d.empId = dg.empId
)
select empId, min(orderDate) as MinDate, max(orderDate) as MaxDate
from dateGroups
where empId = 4
group by empId, groupNum
Basically it first makes a CTE to get the row numbers for each row in date order. Then it makes a recursive CTE that first finds all the groups with no previous sequential entries then adds all subsequent entries to the same group. Finally it takes the records with all the group numbers assigned and groups them by their group number and gets the min and max dates.

TSQL OVER CLAUSE That has no partition by has Order By clause

I am having problems reading code like
SELECT
employeeID as ID,
RANK() OVER (ORDER BY AVG (Salary) DESC) AS Value
FROM Salaries
which supposedly gets the average salary of every employees
My understanding is the code should be
SELECT
employeeID as ID,
RANK() OVER (Partition By employeeID ORDER BY AVG (Salary) DESC) AS Value
FROM Salaries
but the above code works just fine?
First one is not working for me (returning Msg 8120
Column 'Salaries.employeeID' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause), until I add group by employeeID:
SELECT
employeeID as ID,
RANK() OVER (ORDER BY AVG (Salary) DESC) AS Value
FROM Salaries
GROUP BY employeeID
Perhaps, for better understanding, it can be rewritten equivalently as:
;with cte as (
SELECT employeeID, AVG (Salary) as AvgSalary
FROM Salaries
GROUP BY employeeID
)
select employeeID as ID
, RANK() OVER (ORDER BY AvgSalary DESC) as Value
--, AvgSalary
from cte
In this case, average salary by employee is calculated in the CTE, and then query is extended with ranking column Value. Adding partition by employeeID to over clause:
;with cte as (
SELECT employeeID, AVG (Salary) as AvgSalary
FROM Salaries
GROUP BY employeeID
)
select employeeID as ID
, RANK() OVER (partition by employeeID ORDER BY AvgSalary DESC) as Value
--, AvgSalary
from cte
will lead to Value = 1 for every row in the result set (which is not what seem attempted to be achieved), because of rank() will reset numbering to 1 for each distinct employeeID, and employeeID is distinct in every row, since data was aggregated by this column prior to ranking.

Resources