How to use partition in Laravel - mysql

I have a table of employees with their name, dept and salary.
See here
I want to fetch 2 highest salary employees from every dept
This is how I achieved in MySql,
select dept, name, salary from (
select dept, name, salary,
row_number() over (
partition by dept
order by salary desc
) rn
from
employees as T
)as T
where T.rn < 2;
How would I write this with
Laravel Eloquent or Query Builder (Without using raw)?

Related

I am trying to get latest month salary from salary table, where multiple entry of employeeID in salary table for each month

Employee table"
empno,
ename,
sal_id,
emp_id,
salary,
month
SALARY table:
sal_id,
emp_id,
salary,
month
I am trying to make query for getting all employee from employee table +
In salary table, there are multiple entry (may be not) of emp_id.
I want employee list with their latest salary( or last month salary)
My current query is :
SELECT * FROM emp LEFT JOIN salary ON emp.empno = salary.emp_id GROUP BY empno ORDER BY salary.sal_id DESC
But I am getting emp list with first salary, I want with latest salary.
Help me :(
Emploee table
Salary table

You are ordering by sal_id instead you might want to order by a field in salary which represents its month.

SELECT emp.empno, emp.name, (
SELECT sal
FROM salary
WHERE salary.empno = emp.empno
ORDER BY salary.sal_id DESC LIMIT 1
)
FROM emp

I have come with solution after a lot R&D and googling .
It is not possible in single query, we have to apply 2 query
SELECT * FROM emp LEFT JOIN salary ON salary.emp_id=emp.empno WHERE salary.sal_id IN ( SELECT MAX(salary.sal_id) FROM salary GROUP BY salary.emp_id ) GROUP BY salary.sal_id ORDER BY salary.sal_id DESC

How to get second highest salaried employees in mysql

I have an employee table in MySQL with below entries. I need to find all the employees having second highest salaries. In this case, it would be c and d.
id | name | salary
1 | a | 1000
2 | b | 1000
3 | c | 500
4 | d | 500
5 | e | 400
I tried running below query
SELECT name, MAX(salary) FROM employee WHERE salary < (SELECT MAX(salary) from employee);
But this query returns just c as a result. How to get both c and d in result?
I looked at bunch of similar questions posted but none of them mentioned how to get multiple rows for second highest salary.

You can esily get second highest salary from table
SELECT MAX(salary) FROM Employee WHERE Salary NOT IN ( SELECT Max(Salary) FROM Employee);

You can find the second highest salary with:
SELECT salary
FROM employee
GROUP BY salary
ORDER BY salary DESC
LIMIT 1, 1
Then either feed the result of that to another query in the same transaction:
SELECT *
FROM employee
WHERE salary = ?
Or do it as a subquery:
SELECT *
FROM employee
WHERE salary = (
SELECT salary
FROM employee
GROUP BY salary
ORDER BY salary DESC
LIMIT 1, 1
)

In case you want migrate to MSSQL Server :).
SELECT * FROM (
SELECT MAX(salary) T,RANK() OVER (ORDER BY SALARY DESC) AS RankBySalary FROM Employees
GROUP BY SALARY ) TB
WHERE RankBySalary = 3
Or much better:
SELECT * FROM
(
SELECT ID,NAME,SALARY,DENSE_RANK() OVER (ORDER BY SALARY DESC) AS RankBySalary FROM employee
)
TB WHERE RankBySalary = 2

SELECT *
FROM employee one1
WHERE ( N ) = (
SELECT COUNT( one2.salary )
FROM employee one2
WHERE one2.salary > one1.salary
)
Note : N means Nth highest salary
Demo

First find the second highest salary amount then select the rows having that salary.
Query
select * from Employees
where Salary = (
select min(t.salary) from (
select salary
from Employees
group by salary
order by salary desc limit 2
)t
);
SQL Fiddle demo

I suggest that you must first select the 2nd highest salary first and then use the derived table with JOIN on original table. like this:
SELECT
original_record.*
FROM
salary_record AS original_record
JOIN
(SELECT
distinct salary
FROM
salary_record
ORDER BY 1
LIMIT 1,1
) AS derived_record
ON
original_record.salary = derived_record.salary
PS: I have renamed your employee table as salary_record table
Also have a look at Varoon Sahgal's article on Nth highest salary, here: http://www.programmerinterview.com/index.php/database-sql/find-nth-highest-salary-sql/ . The comments-section of this article as well as the article itself has some optimized examples.

select name,max(salary) from employee x where (n-1)=(select count(distinct salary)from employee y where x.salary<y.salary);
Nth max salary

-- IF THIS TABLE EXISTS, DROP IT
DROP TABLE E2;
-- THE FOLLOWING CTE ARRANGES SALARIES IN DECENDING ORDER
WITH copytable(Salary) AS
(
SELECT Emp1.Salary
FROM Employees AS Emp1, Employees AS Emp2
WHERE Emp1.Salary > Emp2.salary
GROUP BY Emp1.Salary
)
-- COPY THE CTE IN A TABLE
SELECT * INTO E2 FROM copytable
-- GIVES THE RANK TO THE SALARY IN DECENDING ORDER
ALTER TABLE E2
ADD RankOfSalary INT IDENTITY(1, 1)
-- TO GET THE LOWEST RANK WHICH WILL BE THE HIGHEST SALARY
DECLARE #rankOfSalary int;
SELECT #rankOfSalary = COUNT(Salary) from E2;
-- SELECTING THE SECOND LARGEST SALARY
DECLARE #SelectSalary INT
SELECT #SelectSalary = Salary from E2 where RankOfSalary = #rankOfSalary - 1;
THIS IS HOW YOU DO IT WITHOUT USING max(), order by, top in sql server
to select second largest salary of employees without using max(), order by, top in sql server
JUST WANTED TO POST THIS :p

You can get it by this:
2nd Largest Salary:
SELECT name, salary
FROM employees
WHERE salary = (SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees))
3rd Largest Salary:
SELECT name, salary
FROM employees
WHERE salary = (SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees)))
Helpful links:
http://www.mysqltutorial.org/select-nth-highest-record-database-table-using-mysql.aspx
http://www.programmerinterview.com/index.php/database-sql/find-nth-highest-salary-sql/
http://www.coderanch.com/t/530503/JDBC/databases/select-Nth-highest-salary-table

try this TO GET BOTH c AND d
SELECT
name, salary
FROM
employee
WHERE salary = (
SELECT salary
FROM employee
GROUP BY salary
ORDER BY salary DESC
LIMIT 1,1
);

Fixing a Query in mysql

I need to display the code of the Department in the School table with the highest total Salary of any department.
So I tried this:
SELECT MAX(Total), dept
FROM (SELECT SUM(salary) AS Total, dept from school group by dept) AS Temp;
Which gives me the correct result; however it shows the field MAX(Total) and I just want the code of the department to show. What should I change?

You can do
SELECT a.dept FROM (
SELECT MAX(Total), dept
FROM (SELECT SUM(salary) AS Total, dept from school group by dept) AS Temp
) AS a;

If you need department code with max salary fund:
SELECT a.dept
FROM (SELECT SUM(salary) AS Total, dept from school group by dept) AS a
ORDER BY a.Total DESC
LIMIT 1;

Find the dept that is being paid the max salary

I'm a student and am working on a MySql assignment given to us for the holidays.
I have a table that looks something like this :
name dept salary
A Sales 100
B Marketing 200
C Sales 800
(Sorry I'm new to stackexchange so I don't know how to display a table.)
The question for the query is : find the dept that is being paid the max salary.
I entered the following query :
SELECT dept
, SUM(salary)
FROM emp
GROUP BY dept
HAVING MAX(SUM(salary));
But I am getting the following error:
'Invalid use of group function error'.

In case salary sum is unique, you can calculate sum for each dept then order records in descending order by aggregated salary and fetch the first record (with the greatest salary):
select dept
, sum(salary) as salary
from tbl
group by dept
order by salary desc
limit 1
In case salary sum can be the same for multiple depts you can calculate salary sum for each dept , then find maximum salary sum the same way as explained above and using a having clause validate if salary sum for each group is equal to the maximum salary sum:
select dept
from tbl
group by dept
having sum(salary) = ( select sum(salary) as salary
from tbl
group by dept
order by salary desc
limit 1)

Find max and second max salary for a employee table MySQL

Suppose that you are given the following simple database table called Employee that has 2 columns named Employee ID and Salary:
Employee
Employee ID Salary
3 200
4 800
7 450
I wish to write a query select max(salary) as max_salary, 2nd_max_salary from employee
then it should return
max_salary 2nd_max_salary
800 450
i know how to find 2nd highest salary
SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )
or to find the nth
SELECT FROM Employee Emp1 WHERE (N-1) = ( SELECT COUNT(DISTINCT(Emp2.Salary)) FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
but i am unable to figureout how to join these 2 results for the desired result

You can just run 2 queries as inner queries to return 2 columns:
select
(SELECT MAX(Salary) FROM Employee) maxsalary,
(SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )) as [2nd_max_salary]
SQL Fiddle Demo

Try like this
SELECT (select max(Salary) from Employee) as MAXinmum),(max(salary) FROM Employee WHERE salary NOT IN (SELECT max(salary)) FROM Employee);
(Or)
Try this, n would be the nth item you would want to return
SELECT DISTINCT(Salary) FROM table ORDER BY Salary DESC LIMIT n,1
In your case
SELECT DISTINCT(column_name) FROM table_name ORDER BY column_name DESC limit 2,1;

Simplest way to fetch second max salary & nth salary
select
DISTINCT(salary)
from employee
order by salary desc
limit 1,1
Note:
limit 0,1 - Top max salary
limit 1,1 - Second max salary
limit 2,1 - Third max salary
limit 3,1 - Fourth max salary

The Best & Easiest solution:-
SELECT
max(salary)
FROM
salary
WHERE
salary < (
SELECT
max(salary)
FROM
salary
);

You can write 2 subqueries like this example
SELECT (select max(Salary) from Employee) as max_id,
(select Salary from Employee order by Salary desc limit 1,1) as max_2nd

Select Distinct sal From emp Order By sal Desc Limit 1,1;
It will take all Distinct sal. And Limit 1,1 means: leaves top one record and print 1 record.

$q="select * from info order by salary desc limit 1,0"; // display highest 2 salary
or
$q="select * from info order by salary desc limit 1,0"; // display 2nd highest salary

I think, It is the simplest way to find MAX and second MAX Salary.You may try this way.
SELECT MAX(Salary) FROM Employee; -- For Maximum Salary.
SELECT MAX(Salary) FROM Employee WHERE Salary < (SELECT MAX(Salary) FROM Employee); -- For Second Maximum Salary

i think that the simple way in oracle is this:
SELECT Salary FROM
(SELECT DISTINCT Salary FROM Employee ORDER BY Salary desc)
WHERE ROWNUM <= 2;

`select max(salary) as first, (select salary from employee order by salary desc limit 1, 1) as second from employee limit 1`
For max salary simply we can use max function, but second max salary we should use sub query. in sub query we can use where condition to check second max salary or simply we can use limit.

You can write SQL query in any of your favorite database e.g. MySQL, Microsoft SQL Server or Oracle. You can also use database specific feature e.g. TOP, LIMIT or ROW_NUMBER to write SQL query, but you must also provide a generic solution which should work on all database. In fact, there are several ways to find second highest salary and you must know a couple of them e.g. in MySQL without using the LIMIT keyword, in SQL Server without using TOP and in Oracle without using RANK and ROWNUM.
Generic SQL query:
SELECT
MAX(salary)
FROM
Employee
WHERE
Salary NOT IN (
SELECT
Max(Salary)
FROM
Employee
);
Another solution which uses sub query instead of NOT IN clause. It uses < operator.
SELECT
MAX(Salary)
FROM
Employee
WHERE
Salary < (
SELECT
Max(Salary)
FROM
Employee
);

This solution will give all employee name and salary who have second highest salary
SELECT name, salary
FROM employee
WHERE salary = (
SELECT
salary
FROM employee AS emp
ORDER BY salary DESC
LIMIT 1,1
);

Find Max salary of an employee
SELECT MAX(Salary) FROM Employee
Find Second Highest Salary
SELECT MAX(Salary) FROM Employee
Where Salary Not In (Select MAX(Salary) FROM Employee)
OR
SELECT MAX(Salary) FROM Employee
WHERE Salary <> (SELECT MAX(Salary) FROM Employee )

This will be the simplest code format :
select max(salary) as 'max_salary',
(select salary from employee order by salary desc limit 1,1) as
'2nd_max_salary'
from employee;
For finding the nth highest salary, syntax is :
select max(salary) as 'max_salary',
(select salary from employee order by salary desc limit n-1,1) as
'nth_max_salary'
from employee;

Not really a nice query but :
SELECT * from (
SELECT max(Salary) from Employee
) as a
LEFT OUTER JOIN
(SELECT MAX(Salary) FROM Employee
WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )) as b
ON 1=1

For unique salaries (i.e. first can't be same as second):
SELECT
MAX( s.salary ) AS max_salary,
(SELECT
MAX( salary )
FROM salaries
WHERE salary <> MAX( s.salary )
ORDER BY salary DESC
LIMIT 1
) AS 2nd_max_salary
FROM salaries s
And also because it's such an unnecessary way to go about solving this problem (Can anyone say 2 rows instead of 2 columns, LOL?)

Try
SELECT
SUBSTRING( GROUP_CONCAT( Salary ), 1 , LOCATE(",", GROUP_CONCAT( Salary ) ) -1 ) AS max_salary,
SUBSTRING( GROUP_CONCAT( Salary ), LOCATE(",", GROUP_CONCAT( Salary ) ) +1 ) AS second_max_salary
FROM
(
SELECT Salary FROM `Employee` ORDER BY Salary DESC LIMIT 0,2
) a
Demo here

For second highest salary, This one work for me:
SELECT salary
FROM employee
WHERE salary
NOT IN (
SELECT MAX( salary )
FROM employee
ORDER BY salary DESC
)
LIMIT 1

This is awesome Query to find the nth Maximum:
For example: -
You want to find salary 8th row Salary, Just Changed the indexed value to 8.
Suppose you have 100 rows with Salary. Now you want to find Max salary for 90th row. Just changed the Indexed Value to 90.
set #n:=0;
select * from (select *, (#n:=#n+1) as indexed from employee order by Salary desc)t where t.indexed = 1;

with Common table expression
With cte as (
SELECT
ROW_NUMBER() Over (Order By Salary Desc) RowNumber,
Max(Salary) Salary
FROM
Employee
Group By Salary
)
Select * from cte where RowNumber = 2

without nested query
select max(e.salary) as max_salary, max(e1.salary) as 2nd_max_salary
from employee as e
left join employee as e1 on e.salary != e1.salary
group by e.salary desc limit 1;

Here change n value according your requirement:
SELECT top 1 amount
FROM (
SELECT DISTINCT top n amount
FROM payment
ORDER BY amount DESC ) AS temp
ORDER BY amount

This should work same :
SELECT MAX(salary) max_salary,
(SELECT MAX(salary)
FROM Employee
WHERE salary NOT IN
(SELECT MAX(salary)
FROM Employee)) 2nd_max_salary
FROM Employee

If we want to find Employee that gets 3nd highest salary then execute this query
SELECT a.employeeid,
a.salary
FROM (SELECT employeeid,
salary,
Dense_rank()
OVER(
ORDER BY salary) AS n
FROM employee) AS a
WHERE n = 3
What do you want

This will work To find the nth maximum number
SELECT
TOP 1 * from (SELECT TOP nth_largest_no * FROM Products Order by price desc) ORDER BY price asc;
For Fifth Largest number
SELECT
TOP 1 * from (SELECT TOP 5 * FROM Products Order by price desc) ORDER BY price asc;

Here is another solution which uses sub query but instead of IN clause it uses < operator
SELECT MAX(Salary) From Employees WHERE Salary < ( SELECT Max(Salary) FROM Employees);

select * from emp where sal =(select max(sal) from emp where eno in(select eno from emp where sal <(select max(sal)from emp )));
try the above code ....
if you want the third max record then add another nested query "select max(sal)from emp" inside the bracket of the last query and give less than operator in front of it.

select * from
Employees where Sal >=
(SELECT
max(Sal)
FROM
Employees
WHERE
Sal < (
SELECT
max(Sal)
FROM
Employees;
));

Max Salary:
select max(salary) from tbl_employee <br><br>
Second Max Salary:
select max(salary) from tbl_employee where salary < ( select max(salary) from tbl_employee);

Try below Query, was working for me to find Nth highest number salary.
Just replace your number with nth_No
Select DISTINCT TOP 1 salary
from
(Select DISTINCT TOP *nth_No* salary
from Employee
ORDER BY Salary DESC)
Result
ORDER BY Salary