Related
I'm trying to understand below query, how its working.
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary
)
Lets say I have 5 distinct salaries and want to get 3rd largest salary. So Inner query will run first and then outer query ?
I'm getting confused how its being done in sql engine. Curious to know. Becasue if its 3rd largest then 3-1 = 2, so that 2 needs to be matched with inner count as well. How inner count is being operated.
Can anyone explain the how its working .. ?
The subquery is correlated subquery, so it conceptually executes once for each row in the outer query (database optimizations left apart).. What it does is count how many employees have a salary greater than the one on the row in the outer query: if there are 2 employee with a higher salary, then you know that the employee on the current row in the outer query has the third highest salary.
Another way to phrase this is to use row_number() for this:
select *
from (
select
e.*,
row_number() over(order by salary desc) rn
from employee e
) t
where rn = 3
Depending on how you want to handle duplicates, dense_rank() might also be an option.
SELECT * FROM (SELECT EMP.ID,RANK() OVER (ORDER BY SALARY DESC) AS NOS FROM EMPLOYEE) T WHERE T.NOS=3
Then from this select the one with any desired rank.
It is easier to understand when you run this query:
select e1.*,
(select count(distinct e2.salary)
from employee e2
where e2.salary > e1.salary) as n
from employee e1
This is my sample table:
create table employee(salary) as (
select * from table(sys.odcinumberlist(1500, 1200, 1400, 1500, 1100)));
so my output is:
SALARY N
---------- ----------
1500 0
1200 2
1400 1
1500 0
1100 3
As you can see that subquery counts, for each row, salaries which are greater than salary in current row. So for instance, for 1400 there is one DISTINCT greater salary (1500). 1500 appears twice in my table, but distinct makes that it is counted once. So 1400 is second in order.
Your query has this count moved to the where part and compared with required value. We have to substract one, because for highest salary there is no higher value, for second salary one row etc.
It's one of the methods used to find such values, newer Oracle versions introduced analytic functions (rank, row_number, dense_rank) which eliminates the need of using subqueries for such purposes. They are faster, more efficient. For your query dense_rank() would be useful.
I need to write a query that will return the third highest salaried employee in the company.
I was trying to accomplish this with subqueries, but could not get the answer. My attempts are below:
select Max(salary)
from employees
where Salary not in
(select Max(salary)
from employees
where Salary not in
(select Max(salary)
from employees));
My thought was that I could use 2 subqueries to elimitate the first and second highest salaries. Then I could simply select the MAX() salary that is remaining. Is this a good option, or is there a better way to achieve this?
The most simple way that should work in any database is to do following:
SELECT * FROM `employee` ORDER BY `salary` DESC LIMIT 1 OFFSET 2;
Which orders employees by salary and then tells db to return a single result (1 in LIMIT) counting from third row in result set (2 in OFFSET). It may be OFFSET 3 if your DB counts result rows from 1 and not from 0.
This example should work in MySQL and PostgreSQL.
Edit:
But there's a catch if you only want the 3rd highest DISTINCT salary. Than you should add the DISTINCT keyword.
In case of salary list: 100, 90, 90, 80, 70.
In the above query it will produce the 3rd highest salary which is 90. But if you mean the 3rd distinct which is 80 than you should use
SELECT DISTINCT `salary` FROM `employee` ORDER BY `salary` DESC LIMIT 1 OFFSET 2;
But there's a catch, this will return you only 1 column which is Salary, because in order to operate the distinction operation, DISTINCT can only operate on a specific set of columns.
This means we should add another wrapping query to extract the employees(There can be multiple) that matches that result. Thus I added LIMIT 1 at the end.
SELECT *
FROM `employee`
WHERE
`Salary` = (SELECT DISTINCT `Salary`
FROM `employee`
ORDER BY `salary` DESC
LIMIT 1 OFFSET 2
)
LIMIT 1;
Examples can be found HERE
You can get the third highest salary by using limit , by using TOP keyword and sub-query
TOP keyword
SELECT TOP 1 salary
FROM
(SELECT TOP 3 salary
FROM Table_Name
ORDER BY salary DESC) AS Comp
ORDER BY salary ASC
limit
SELECT salary
FROM Table_Name
ORDER BY salary DESC
LIMIT 2, 1
by subquery
SELECT salary
FROM
(SELECT salary
FROM Table_Name
ORDER BY salary DESC
LIMIT 3) AS Comp
ORDER BY salary
LIMIT 1;
I think anyone of these help you.
You may try (if MySQL):
SELECT salary FROM employee ORDER BY salary DESC LIMIT 2, 1;
This query returns one row after skipping two rows.
You may also want to return distinct salary. For example, if you have 20,20,10 and 5 then 5 is the third highest salary. To do so, add DISTINCT to the above query:
SELECT DISTINCT salary FROM employee ORDER BY salary DESC LIMIT 2, 1;
SELECT Max(salary)
FROM employee
WHERE salary < (SELECT Max(salary)
FROM employee
WHERE salary NOT IN(SELECT Max(salary)
FROM employee))
hope this helped you
If SQL Server this could work
SELECT TOP (1) * FROM
(SELECT TOP (3) salary FROM employees ORDER BY salary DESC) T
ORDER BY salary ASC
As for your number of subqueries question goes it depends on your language. Check this for more information
Is there a nesting limit for correlated subqueries in Oracle?
SELECT id
FROM tablename
ORDER BY id DESC
LIMIT 2 , 1
This is only for get 3rd highest value .
You may use this for all employee with 3rd highest salary:
SELECT * FROM `employee` WHERE salary = (
SELECT DISTINCT(`salary`) FROM `employee` ORDER BY `salary` DESC LIMIT 1 OFFSET 2
);
Some DBMS's don't allow you to run several nested queries. Here is a solution that only uses 1 nested query:
SELECT salary
FROM
(
SELECT salary
FROM employees
ORDER BY salary
LIMIT 3
) as TBL1
ORDER BY salary DESC
LIMIT 1;
It should give you the desired result. It first finds the 3 largest salaries, then selects the smallest of the three (or the third one if they are equal). Here is an SQLFiddle
I found a very good explanation in
http://www.programmerinterview.com/index.php/database-sql/find-nth-highest-salary-sql/
This query should give nth highest salary
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
SELECT MAX(salary) FROM employees GROUP BY salary ORDER BY salary DESC LIMIT 1 OFFSET 2;
SELECT * FROM employee ORDER BY salary DESC LIMIT 1 OFFSET 2;
You can use nested query to get that, like below one is explained for the third max salary. Every nested salary is giving you the highest one with the filtered where result and at the end it will return you exact 3rd highest salary irrespective of number of records for the same salary.
select * from users where salary < (select max(salary) from users where salary < (select max(salary) from users)) order by salary desc limit 1
Below query will give accurate answer. Follow and give me comments:
select top 1 salary from (
select DISTINCT top 3 salary from Table(table name) order by salary ) as comp
order by personid salary
you can get any order for salary with that:
select * from
(
select salary,row_Number() over (order by salary DESC ) RN
FROM employees
)s
where RN = 3
-- put RN equal to any number of orders.
--for your question put 3
You can find Nth highest salary by making use of just one single query which is very simple to understand:-
select salary from employees e1 where N-1=(select count(distinct
salary) from employees e2 where e2.salary>e1.salary);
Here Replace "N" with number(1,2,3,4,5...).This query work properly even when where salaries are duplicate. The simple idea behind this query is that the inner subquery
count how many salaries are greater then (N-1). When we get the count then the cursor will point to that row which is N and it simply returns the salary present in that row.
SELECT salary FROM employees e1
WHERE N-1 = (SELECT COUNT(DISTINCT salary) FROM employees e2
WHERE e2.salary > e1.salary)
Here, I have solved it with a correlated nested query. It is a generalized Query so if you want to print 4th, 5th, or any number of highest salary it will work perfectly even if there are any duplicate salaries.
So, what you have to do is simply change the N value here. So, in your case, it will be,
SELECT salary FROM employees e1
WHERE 3-1 = (SELECT COUNT(DISTINCT salary) FROM employees e2
WHERE e2.salary > e1.salary)
Note that the third highest salary may be the same the the first highest salary so your current approach wouldn't work.
I would do order the employees by salary and apply a LIMIT 3 at the end of the SQL query. You'll then have the top three of highest salaries and, thus, you also have the third highest salary (if there is one, a company may have two employees and then you wouldn't have a third highest salary).
For me this query work fine in Mysql
it will return third max salary from table
SELECT salary FROM users ORDER BY salary DESC LIMIT 1 OFFSET 2;
or
SELECT salary FROM users ORDER BY salary DESC LIMIT 2,1;
select min (salary) from Employee where Salary in (Select Top 3 Salary from Employee order by Salary desc)
SELECT TOP 1 BILL_AMT Bill_Amt FROM ( SELECT DISTINCT TOP 3 NH_BL_BILL.BILL_AMT FROM NH_BL_BILL ORDER BY BILL_AMT DESC) A
ORDER BY BILL_AMT ASC
SELECT DISTINCT MAX(salary) AS max
FROM STAFF
WHERE salary IN
(SELECT salary
FROM STAFF
WHERE salary<(SELECT MAX(salary) AS maxima
FROM STAFF
WHERE salary<
(SELECT MAX(salary) AS maxima
FROM STAFF))
GROUP BY salary);
I have tried other ways they are not right. This one works.
We can find the Top nth Salary with this Query.
WITH EMPCTE AS (
SELECT E.*, DENSE_RANK() OVER(ORDER BY SALARY DESC) AS DENSERANK
FROM EMPLOYEES E
)
SELECT * FROM EMPCTE WHERE DENSERANK=&NUM
for oracle it goes like this:
select salary from employee where rownnum<=3 order by salary desc
minus
select salary from employee where rownnum<=2 order by salary desc;
The SQL-Server implementation of this will be:
SELECT SALARY FROM EMPLOYEES OFFSET 2 ROWS FETCH NEXT 1 ROWS ONLY
This is a MYSQL query.
Explanation: The subquery returns top 3 salaries. From the returned result, we select the minimum salary, which is the 3rd highest salary.
SELECT MIN(Salary)
FROM (
SELECT Salary
FROM Employees
ORDER BY Salary DESC
LIMIT 3
) AS TopThreeSalary;
in Sql Query you can get nth highest salary
select * from(
select empname, sal, dense_rank()
over(order by sal desc)r from Employee)
where r=&n;
To find to the 2nd highest sal set n = 2
To find 3rd highest sal set n = 3 and so on.
This works fine with Oracle db.
select SAL from ( SELECT DISTINCT SAL FROM EMP ORDER BY SAL DESC FETCH FIRST 3 ROWS ONLY ) ORDER BY SAL ASC FETCH FIRST 1 ROWS ONLY
SELECT *
FROM maintable_B7E8K
order by Salary
desc limit 1 offset 2;
--Oracle SQL
with temp as (
select distinct salary from HR.EMPLOYEES
order by SALARY desc
)
select min(temp.salary) from temp
where rownum <= 3;
SELECT * FROM(
SELECT salary, DENSE_RANK()
OVER(ORDER BY salary DESC)r FROM Employee)
WHERE r=&n;
To find the 3rd highest salary set n = 3
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
HI,
Can u tell me the syntax of the SQL command which gives as output the second highest salary from a range of salaries stored in the employee table.
A description of the SQL commnd will be welcomed...
Please help!!!
select min(salary) from
(select top 2 salary from SalariesTable order by salary desc)
as ax
This should work:
select * from (
select t.*, dense_rank() over (order by salary desc) rnk from employee t
) a
where rnk = 2;
This returns the second highest salary.
dense_rank() over is a window function, and it gives you the rank of a specific row within the specified set. It is standard SQL, as defined in SQL:2003.
Window functions are awesome in general, they simplyfy lots of difficult queries.
Slightly different solution:
This is identical except that returns the highest salary when there is a tie for number 1:
select * from (
select t.*, row_number() over (order by salary desc) rnk from employee t
) a
where rnk = 2;
Updated: Changed rank to dense_rank and added second solution. Thanks, IanC!
with tempTable as(
select top 2 max(salary) as MaxSalary from employee order by salary desc
) select top 1 MaxSalary from tempTable
description:
select the top 2 maximum salaries
order them by desc order ( so the 2nd highest salary is now at the top)
select the top 1 from that
another approach:
select top 1 MaxSalary from (
select top 2 max(salary) as MaxSalary from employee order by salary desc
)
Here's some sample code, with proof of concept:
declare #t table (
Salary int
)
insert into #t values (100)
insert into #t values (900)
insert into #t values (900)
insert into #t values (400)
insert into #t values (300)
insert into #t values (200)
;WITH tbl AS (
select t.Salary, DENSE_RANK() OVER (order by t.Salary DESC) AS Rnk
from #t AS t
)
SELECT *
FROM tbl
WHERE Rnk = 2
DENSE_RANK is mandatory (change to RANK & you'll see).
You'll also see why any SELECT TOP 2 queries won't work (without a DISTINCT anyway).
You don't specify the actual SQL product you're using, and the query language varies among products. However, something like this should get you started:
SELECT salary FROM employees E1
WHERE 1 = (SELECT COUNT(*) FROM employee E2 WHERE E2.salary > E1.salary)
(thanks to fredt for the correction).
Alternatively (and faster in terms of performance) would be
SELECT TOP 2 salary FROM employees ORDER BY salary DESC
and then skipping the first returned row.
An alternative (tested):
select Min(Salary) from (
select distinct TOP (2) salary from employees order by salary DESC) AS T
This will work on any platform, is clean, and caters for the possibility of multiple tied #1 salaries.
Select top 1 * from employee where empID in (select top 2 (empID) from employee order by salary DESC) ORDER BY salary ASC
Explanation:
select top 2 (empID) from employee order by salary DESC would give the two records for which Salary is top and then the whole query would sort it these two records in ASCENDING order and then list out the one with the lowest salary among the two.
EX. let the salaries of employees be 100, 99, 98,,50.
Query 1 would return the emp ID of the persons with sal 100 and 99
Whole query would return all data related to the person having salary 99.
SELECT Salary,EmpName
FROM
(
SELECT Salary,EmpName,ROW_NUMBER() OVER(ORDER BY Salary) As Rank
FROM EMPLOYEE
) A
WHERE A.Rank=n;
where n is the number of highest salary u r requesting the table.
Ucan also use DenseRank() function in place of ROW_NUMBER().
Thanks,
Suresh
An easier way..
select MAX(salary) as SecondMax from test where salary !=(select MAX(salary) from test)
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