I'm a beginner to oracle. In recent search I've seen WHERE N-1,3-2 ..so on.
How does it work in searching data?
This is my code attemtp so far:
SELECT name, salary
FROM #Employee e1
WHERE N-1 = (SELECT COUNT(DISTINCT salary) FROM #Employee e2
WHERE e2.salary > e1.salary)
It's a classic( and pretty old ) SQL query to get nth highest salary. I assume It is no longer used( I haven't seen ) in any production codes, but could be a favourite question among the interviewers.
The N you are referring to is not a column or some unknown entity but a placeholder which should translate to a valid integer or bind argument in the working query. It is a correlated subquery, a subquery that is evaluated once for each row processed by the outer query. The way it works is that it takes a count of distinct list of salary values from employees that have a salary greater than each one of employees coming from the outer query and restricts the result where that count is equal to N-1.Which means you get those rows with nth highest salaries.
A more commonly used way to do this would be to use analytic function dense_rank() ( or rank depending on your need ). Read the documentation for these functions in case you aren't aware of them.
SELECT first_name,
salary
FROM (
SELECT e.*,
dense_rank() OVER(
ORDER BY salary desc
) rn
FROM employees e
)
WHERE rn = 6; -- ( n = 6 )
In Oracle 12c and higher versions, even though the above query works, a handy option to use is the FETCH..FIRST syntax.
SELECT *
FROM employees
ORDER BY salary DESC OFFSET 6 ROWS FETCH FIRST 1 ROWS WITH TIES; --n=6
In my Employee table, I wanted to find the 3rd highest salary. Someone provided me with the following query to do this:
SELECT *
FROM employee C1
WHERE 3 = (SELECT Count(DISTINCT( C2.salary ))
FROM employee C2
WHERE C2.salary >= C1.salary)
This query works, but I don't how it works. What kind of query is this?
As others have said, this type of query is called a correlated sub-query. It's a sub-query because there is a query within a query and it's correlated because the inner query references the outer query in its definition.
Consider the inner query:
SELECT Count(DISTINCT( C2.salary ))
FROM employee C2
WHERE C2.salary >= C1.salary
Conceptually, this inner query will be evaluated once for every row produced by the outer query before the WHERE clause is applied, basically once for every row in employee. It will produce a single value, the count of rows from employee where the salary is less than the salary of the outer row.
The outer query will only return records where the value produced by the inner query is exactly 3. Assuming unique salary values, there is only one row from the employee table where there will be exactly 3 records with a salary value greater than or equal to it (the one row) and that one row is necessarily the third-highest salary value.
It's clever, but unnecessarily weird and probably not as optimal as something more straightforward.
Maybe a better solution would have been
SELECT TOP 1 *
FROM (
SELECT TOP 3 * FROM employee ORDER BY Salary DESC
) t
ORDER BY Salary ASC
Easier to read and more efficient than a correlated sub-query.
You can also use Dense_Rank to rank the salaries greatest to least, then select the ones that are ranked 3rd. This will also prevent you from getting the wrong salary if the top 2 are identical like the other answers above mine are doing. This has a better looking execution plan than the Distinct count one also
SELECT *
FROM (
SELECT *,
DENSE_RANK() OVER (ORDER BY Salary DESC) salary_rank
FROM employee e
) t
WHERE salary_rank = 3
Could also rewrite this with a common table expression.
WITH top_three
AS
(
SELECT TOP 3 * FROM employee ORDER BY Salary DESC
)
SELECT TOP 1 *
FROM top_three
ORDER BY Salary ASC;
Or, if you need to look for other ranks in this you can use row_number().
WITH ranked
AS
(
SELECT rank = ROW_NUMBER()OVER(ORDER BY Salary DESC), *
FROM employee
ORDER BY Salary DESC
)
SELECT *
FROM ranked
WHERE rank = #whatever_rank_you_want;
I have a table in which data is something like this:
Name Salary
Tom 10000
John 20000
Ram 20000
Danny 15000
Sandy 14000
Riddle 15000
I can find 2nd highest salary using cte;
with cte
as
(
select ROW_NUMBER() over (order by Salary desc) as r,
* from Employee e
)
select * from cte where r=2
But this gives the result 'Ram' with 20000 salary. What I would like returned is every record for people with the nth-ranking salary. For instance, if I'm looking for n=2, the result would be:
Danny 15000
Riddle 15000
How do I modify the query to achieve this?
Use DENSE_RANK():
;WITH cte AS
(
SELECT DENSE_RANK() OVER (ORDER BY Salary DESC) AS r, *
FROM Employee e
)
SELECT *
FROM cte
WHERE r = 2
SQL Fiddle
I posted about this on my blog:
https://fullparam.wordpress.com/2015/03/31/sql-select-nth-rank-of-something-three-approaches/
Correlated subquery:
SELECT FirstName, LastName, BaseRate
FROM DimEmployee e
WHERE (SELECT COUNT(DISTINCT BaseRate)
FROM DimEmployee p WHERE e.BaseRate=p.BaseRate) = 4
Why is this a good answer? It’s not really but this will work on any SQL implementation.
It’s fairly slow, it will do a lot of look ups. The subquery is evaluated every time a row is processed by the outer query. This query uses dense ranking and can return multiple rows.
Double Order By with TOP statement:
SELECT TOP 1 FirstName, LastName, BaseRate
FROM ( SELECT TOP 4 FirstName, LastName, BaseRate
FROM DimEmployee ORDER BY BaseRate DESC) AS MyTable
ORDER BY BaseRate ASC;
Why is this a good answer? Because it is an easy syntax to remember.
Let’s look at the subquery, which returns the N highest salaries in the DimEmployee table in descending order. Then, the outer query will re-order those values in ascending (default) order, this means the Nth highest salary will now be the topmost salary. Keep in mind that the TOP statement is MS SQL server specific. MySQL would use LIMIT 1 for instance. In addition this solution cannot do DENSE ranking and only returns one row even if two employees share the same BaseRate.
Use Windowing function:
SELECT FirstName, LastName, BaseRate
FROM (SELECT FirstName, LastName, BaseRate, DENSE_RANK() OVER (ORDER BY BaseRate DESC) Ranking
FROM DimEmployee) AS MyTable
WHERE Ranking = 4
Why is this a good answer? Because it performs the best – performance is king. The Syntax is also ANSI SQL however of the “Big 3” only Oracle and MS are using it. In addition you can chose to use ROW_NUMBER, DENSE_RANK or regular RANK.
Try this query:
WITH
cte AS
(select DENSE_RANK() over (ORDER BY Salary desc) AS R, e.*
from Employee e)
SELECT * From cte
WHERE R =2
Here is a SQLFiddle
u can get the 2nd highest salary by using CO-RELATED Query as follows:
select name from employee e1 --outer query
where 2= --2nd highest salary
(select COUNT(e2.salary) --inner query
from employee e2 where e2.salary >= e1.salary)
The DISTINCT keyword might help you here.
SELECT Salary FROM Employee order by Salary desc
will return
20000
20000
15000
15000
14000
10000
but add DISTINCT
SELECT DISTINCT Salary FROM Employee order by Salary desc
will return
20000
15000
14000
10000
So which do you want, the third highest person by salary (which isn't guaranteed to always be the same, as two people might share the 15,000 salary), or the people who have the third highest salary (which in this case, is only one person who has the 14,000 salary, but that's no guarantee.
How to find third or nth maximum salary from salary table(EmpID, EmpName, EmpSalary) in optimized way?
Row Number :
SELECT Salary,EmpName
FROM
(
SELECT Salary,EmpName,ROW_NUMBER() OVER(ORDER BY Salary) As RowNum
FROM EMPLOYEE
) As A
WHERE A.RowNum IN (2,3)
Sub Query :
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary
)
Top Keyword :
SELECT TOP 1 salary
FROM (
SELECT DISTINCT TOP n salary
FROM employee
ORDER BY salary DESC
) a
ORDER BY salary
Use ROW_NUMBER(if you want a single) or DENSE_RANK(for all related rows):
WITH CTE AS
(
SELECT EmpID, EmpName, EmpSalary,
RN = ROW_NUMBER() OVER (ORDER BY EmpSalary DESC)
FROM dbo.Salary
)
SELECT EmpID, EmpName, EmpSalary
FROM CTE
WHERE RN = #NthRow
Try this
SELECT TOP 1 salary FROM (
SELECT TOP 3 salary
FROM employees
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
For 3 you can replace any value...
If you want optimize way means use TOP Keyword, So the nth max and min salaries query as follows but the queries look like a tricky as in reverse order by using aggregate function names:
N maximum salary:
SELECT MIN(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP N EmpSalary FROM Salary ORDER BY EmpSalary DESC)
for Ex: 3 maximum salary:
SELECT MIN(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP 3 EmpSalary FROM Salary ORDER BY EmpSalary DESC)
N minimum salary:
SELECT MAX(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP N EmpSalary FROM Salary ORDER BY EmpSalary ASC)
for Ex: 3 minimum salary:
SELECT MAX(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP 3 EmpSalary FROM Salary ORDER BY EmpSalary ASC)
Too simple if you use the sub query!
SELECT MIN(EmpSalary) from (
SELECT EmpSalary from Employee ORDER BY EmpSalary DESC LIMIT 3
);
You can here just change the nth value after the LIMIT constraint.
Here in this the Sub query Select EmpSalary from Employee Order by EmpSalary DESC Limit 3; would return the top 3 salaries of the Employees. Out of the result we will choose the Minimum salary using MIN command to get the 3rd TOP salary of the employee.
Replace N with your Max Number
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
Explanation
The query above can be quite confusing if you have not seen anything like it before – the inner query is what’s called a correlated sub-query because the inner query (the subquery) uses a value from the outer query (in this case the Emp1 table) in it’s WHERE clause.
And Source
Third or nth maximum salary from salary table without using subquery
select salary from salary
ORDER BY salary DESC
OFFSET N-1 ROWS
FETCH NEXT 1 ROWS ONLY
For 3rd highest salary put 2 in place of N-1
SELECT Salary,EmpName
FROM
(
SELECT Salary,EmpName,DENSE_RANK() OVER(ORDER BY Salary DESC) Rno from EMPLOYEE
) tbl
WHERE Rno=3
SELECT EmpSalary
FROM salary_table
GROUP BY EmpSalary
ORDER BY EmpSalary DESC LIMIT n-1, 1;
Refer following query for getting nth highest salary. By this way you get nth highest salary in MYSQL. If you want get nth lowest salary only you need to replace DESC by ASC in the query.
Method 1:
SELECT TOP 1 salary FROM (
SELECT TOP 3 salary
FROM employees
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
Method 2:
Select EmpName,salary from
(
select EmpName,salary ,Row_Number() over(order by salary desc) as rowid
from EmpTbl)
as a where rowid=3
In 2008 we can use ROW_NUMBER() OVER (ORDER BY EmpSalary DESC) to get a rank without ties that we can use.
For example we can get the 8th highest this way, or change #N to something else or use it as a parameter in a function if you like.
DECLARE #N INT = 8;
WITH rankedSalaries AS
(
SELECT
EmpID
,EmpName
,EmpSalary,
,RN = ROW_NUMBER() OVER (ORDER BY EmpSalary DESC)
FROM salary
)
SELECT
EmpID
,EmpName
,EmpSalary
FROM rankedSalaries
WHERE RN = #N;
In SQL Server 2012 as you might know this is performed more intuitively using LAG().
Answering this question from the point of view of SQL Server as this is posted in the SQL Server section.
There many approaches of getting Nth salary and we can classify these approaches in two sections one using ANSI SQL approach and other using TSQL approach. You can also check out this find nth highest salary youtube video which shows things practically. Let’s try to cover three ways of writing this SQL.
Approach number 1: - ANSI SQL: - Using Simple order by and top keyword.
Approach number 2: - ANSI SQL: - Using Co-related subqueries.
Approach number 3: - TSQL: - using Fetch Next
Approach number 1: - Using simple order by and top.
In this approach we will using combination of order by and top keyword. We can divide our thinking process in to 4 steps: -
Step 1: - Descending :- Whatever data we have first make it descending by using order by clause.
Step 2:- Then use TOP keyword and select TOP N. Where N stands for which highest salary rank you want.
Step 3: - Ascending: - Make the data ascending.
Step 4:- Select top 1 .There you are done.
So, if you put down the above 4 logical steps in SQL it comes up something as shown below.
Below is the text of SQL in case you want to execute and test the same.
select top 1 * from (select top 2 EmployeeSalary from tblEmployee
order by EmployeeSalary desc) as innerquery order by EmployeeSalary
asc
Parameterization issue of Approach number 1
One of the biggest issues of Approach number 1 is “PARAMETERIZATION”.
If you want to wrap up the above SQL in to a stored procedure and give input which top salary you want as a parameter, it would be difficult by Approach number 1.
One of the things you can do with Approach number 1 is make it a dynamic SQL but that would not be an elegant solution. Let’s check out Approach number 2 which is an ANSI SQL approach.
Approach number 2: - Using Co-related subqueries.
Below is how co-related subquery solution will look like. In case you are new to Co-related subquery. Co-related subquery is a query which a query inside query. The outer query first evaluates, sends the record to the inner query, inner query then evaluates and sends it to the outer query.
“3” in the query is the top salary we want to find out.
Select E1.EmployeeSalary from tblEmployee as E1 where 3=(Select
count(*) from tblEmployee as E2 Where
E2.EmployeeSalary>=E1.EmployeeSalary)
So in the above query we have an outer query:-
Select E1.EmployeeSalary from tblEmployee as E1
and inner query is in the where clause. Watch those BOLD’s which indicate how the outer table alias is referred in the where clause which makes co-related evaluate inner and outer query to and fro: -
where 3=(Select count(*) from tblEmployee as E2 Where
E2.EmployeeSalary>=E1.EmployeeSalary)
So now let’s say you have records like 3000, 4000 ,1000 and 100 so below will be the steps: -
First 3000 will be send to the inner query.
Inner query will now check how many record values are greater than or equal to 3000. If the number of record counts is not equal, it will take next value which is 4000. Now for 3000 there are only 2 values which is greater than or equal, 3000 and 4000. So, Is number record count 2>-=3? .NO, so it takes second value which is 4000.
Again for 4000 how many record values are greater than or equal. If the number of record count is not equal, it will take next value which is 1000.
Now 1000 has 3 records more or equal than 1000, (3000,4000 and 1000 himself). This is where co-related stops and exits and gives the final output.
Approach number 3: - TSQL fetch and Next.
Third approach is by using TSQL. By using Fetch and Next, we can get the Nth highest easily.
But please do note, TSQL code will not work for other databases we will need to rewrite the whole code again.
It would be a three-step process:-
Step 1 Distinct and Order by descending: - First apply distinct and order by which made the salaries descending as well as weed off the duplicates.
Step 2 Use Offset: - Use TSQL Offset and get the top N-1 rows. Where N is the highest salary we want to get. Offset takes the number of rows specified, leaving the other rows. Why (N-1) because it starts from zero.
Step 3 Use Fetch: - Use fetch and get the first row. That row has the highest salary.
The SQL looks something as shown below.
Performance comparison
Below is the SQL plan for performance comparison.
Below is the plan for top and order by.
Below is the plan for co-related queries. You can see the number of operators are quiet high in numbers. So surely co-related would perform bad for huge data.
Below is TSQL query plan which is better than cor-related.
So, summing up we can compare more holistically as given in the below table.
declare #maxNthSal as nvarchar(20)
SELECT TOP 3 #maxNthSal=GRN_NAME FROM GRN_HDR ORDER BY GRN_NAME DESC
print #maxNthSal
To get third highest value from table
SELECT * FROM tableName ORDER BY columnName DESC LIMIT 2, 1
This is one of the popular question in any SQL interview. I am going to write down different queries to find out the nth highest value of a column.
I have created a table named “Emloyee” by running the below script.
CREATE TABLE Employee([Eid] [float] NULL,[Ename] [nvarchar](255) NULL,[Basic_Sal] [float] NULL)
Now I am going to insert 8 rows into this table by running below insert statement.
insert into Employee values(1,'Neeraj',45000)
insert into Employee values(2,'Ankit',5000)
insert into Employee values(3,'Akshay',6000)
insert into Employee values(4,'Ramesh',7600)
insert into Employee values(5,'Vikas',4000)
insert into Employee values(7,'Neha',8500)
insert into Employee values(8,'Shivika',4500)
insert into Employee values(9,'Tarun',9500)
Now we will find out 3rd highest Basic_sal from the above table using different queries.
I have run the below query in management studio and below is the result.
select * from Employee order by Basic_Sal desc
We can see in the above image that 3rd highest Basic Salary would be 8500. I am writing 3 different ways of doing the same. By running all three mentioned below queries we will get same result i.e. 8500.
First Way: - Using row number function
select Ename,Basic_sal
from(
select Ename,Basic_Sal,ROW_NUMBER() over (order by Basic_Sal desc) as rowid from Employee
)A
where rowid=2
Select TOP 1 Salary as '3rd Highest Salary' from (SELECT DISTINCT TOP 3 Salary from Employee ORDER BY Salary DESC) a ORDER BY Salary ASC;
I am showing 3rd highest salary
SELECT MIN(COLUMN_NAME)
FROM (
SELECT DISTINCT TOP 3 COLUMN_NAME
FROM TABLE_NAME
ORDER BY
COLUMN_NAME DESC
) AS 'COLUMN_NAME'
--nth highest salary
select *
from (select lstName, salary, row_number() over( order by salary desc) as rn
from employee) tmp
where rn = 2
--(nth -1) highest salary
select *
from employee e1
where 1 = (select count(distinct salary)
from employee e2
where e2.Salary > e1.Salary )
Optimized way: Instead of subquery just use limit.
select distinct salary from employee order by salary desc limit nth, 1;
See limit syntax here http://www.mysqltutorial.org/mysql-limit.aspx
By subquery:
SELECT salary from
(SELECT rownum ID, EmpSalary salary from
(SELECT DISTINCT EmpSalary from salary_table order by EmpSalary DESC)
where ID = nth)
Try this Query
SELECT DISTINCT salary
FROM emp E WHERE
&no =(SELECT COUNT(DISTINCT salary)
FROM emp WHERE E.salary <= salary)
Put n= which value you want
set #n = $n
SELECT a.* FROM ( select a.* , #rn = #rn+1 from EMPLOYEE order by a.EmpSalary desc ) As a where rn = #n
MySQL tested solution, assume N = 4:
select min(CustomerID) from (SELECT distinct CustomerID FROM Customers order by CustomerID desc LIMIT 4) as A;
Another example:
select min(country) from (SELECT distinct country FROM Customers order by country desc limit 3);
Try this code :-
SELECT *
FROM one one1
WHERE ( n ) = ( SELECT COUNT( one2.salary )
FROM one one2
WHERE one2.salary >= one1.salary
)
Find Nth highest salary from a table. Here is a way to do this task using dense_rank() function.
select linkorder from u_links
select max(linkorder) from u_links
select max(linkorder) from u_links where linkorder < (select max(linkorder) from u_links)
select top 1 linkorder
from ( select distinct top 2 linkorder from u_links order by linkorder desc) tmp
order by linkorder asc
DENSE_RANK :
1. DENSE_RANK computes the rank of a row in an ordered group of rows and returns the rank as a NUMBER. The ranks are consecutive integers beginning with 1.
2. This function accepts arguments as any numeric data type and returns NUMBER.
3. As an analytic function, DENSE_RANK computes the rank of each row returned from a query with respect to the other rows, based on the values of the value_exprs in the order_by_clause.
4. In the above query the rank is returned based on sal of the employee table. In case of tie, it assigns equal rank to all the rows.
WITH result AS (
SELECT linkorder ,DENSE_RANK() OVER ( ORDER BY linkorder DESC ) AS DanseRank
FROM u_links )
SELECT TOP 1 linkorder FROM result WHERE DanseRank = 5
In SQL Server 2012+, OFFSET...FETCH would be an efficient way to achieve this:
DECLARE #N AS INT;
SET #N = 3;
SELECT
EmpSalary
FROM
dbo.Salary
ORDER BY
EmpSalary DESC
OFFSET (#N-1) ROWS
FETCH NEXT 1 ROWS ONLY
select * from employee order by salary desc;
+------+------+------+-----------+
| id | name | age | salary |
+------+------+------+-----------+
| 5 | AJ | 20 | 100000.00 |
| 4 | Ajay | 25 | 80000.00 |
| 2 | ASM | 28 | 50000.00 |
| 3 | AM | 22 | 50000.00 |
| 1 | AJ | 24 | 30000.00 |
| 6 | Riu | 20 | 20000.00 |
+------+------+------+-----------+
select distinct salary from employee e1 where (n) = (select count( distinct(salary) ) from employee e2 where e1.salary<=e2.salary);
Replace n with the nth highest salary as number.
SELECT TOP 1 salary FROM ( SELECT TOP n salary FROM employees ORDER BY salary DESC Group By salary ) AS emp ORDER BY salary ASC
(where n for nth maximum salary)
Just change the inner query value: E.g Select Top (2)* from Student_Info order by ClassID desc
Use for both problem:
Select Top (1)* from
(
Select Top (1)* from Student_Info order by ClassID desc
) as wsdwe
order by ClassID
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