Im using Oracle-Apex
I have a table with names and salaries. I want to get the name with the highest salary using MAX(salary).
So the query is like this:
SELECT NAME FROM EMPLOYEE
GROUP BY NAME
HAVING MAX(SALARY) = SALARY;
This doesnt work, error ORA-00979: not a GROUP BY expression, appears. So i use this to stop that error:
SELECT NAME FROM EMPLOYEE
GROUP BY NAME, SALARY
HAVING MAX(SALARY) = SALARY;
And it groups every different salary in one row and returns the max salary in each row, since every salary is different, it returns every row.
How do i group everything in one single big group without modyfing the tables? I mean i want this to work:
SELECT NAME FROM EMPLOYEE
WHERE MAX(SALARY) = SALARY;
But with having. Its simple really, but i cant find the way.
use subquery
SELECT NAME FROM EMPLOYEE
where salary = (select max( salary) from EMPLOYEE)
You need a WHERE clause and compare the salary with the max salary of the table:
SELECT NAME FROM EMPLOYEE
WHERE SALARY = (SELECT MAX(SALARY) FROM EMPLOYEE);
You can get the name of person who has max salary using following nested query.
select NAME
from EMPLOYEE
where SALARY= ( select max(SALARY)
from EMPLOYEE )
Related
I have written the following SQL commands for retrieving data from a table called Employee.
I was able to get the highest/maximum salary as well as the second highest/maximum salary, but I am having difficulty writing the same when the whole record needs to be returned.
Select all the employees.
SELECT * FROM Employee
Return the highest salary. This returns the maximum salary amount which is 90000
SELECT MAX(salary) FROM Employee
Return the employee record with the highest salary. This returns all the records of the person/people with their salary being the maximum salary which is 90000 that is only John Henry in this case.
SELECT *
FROM Employee
WHERE salary = ( SELECT MAX(salary) FROM Employee )
Return every other employee record; i.e. everyone except the one with the highest salary.
SELECT *
FROM Employee
WHERE salary != ( SELECT MAX(salary) FROM Employee )
Return the second highest salary. This returns the second maximum salary amount which is 85000
SELECT MAX(salary) FROM Employee
WHERE salary != ( SELECT MAX(salary) FROM Employee )
Return the employee record with the second highest salary. This returns all the records of the person/people with their salary being the second maximum salary which is 85000 that is only Michael Greenback in this case.
I am stuck in this... I tried using HAVING as an extra condition, but however I arrange it to specify a condition, I get syntax errors. How do I do this?
Window functions are the built-in functionality to do this. In particular, dense_rank():
select e.*
from (select e.*, dense_rank() over (order by salary desc) as seqnum
from employee e
) e
where seqnum = 2;
This is a SQL Server solution but can be easily re-written using limit N instead of top N. If your database doesn't support window functions, you could try this. If it does, Gordon's solution is the way to go.
with cte as
(select max(salary) as second_max
from employee
where salary in (select distinct top 2 salary -- top n is what makes this scalable
from employee
order by salary asc)) -- get top 2 salaries and sort them
select *
from employee
where salary = (select second_max from cte);
employee table
empid(integer)
emp_name(varchar)
salary(integer)
location(varchar)
display employees who have same salary.
how to solve this?
You can use the below query to get the results:
Since you need same salary hence i have grouped out the result with respect to salary and then used a listagg function to retrieve all the employee names with a delimiter.
select salary,listagg(emp_name,' | ') within group (order by emp_name) "Employees"
from employee
group by salary;
Select * from employee where salary is not null
I am trying to find the number of employees in a table that earn exactly the maximum salary of all the employees in the table called tblPerson.
Select Max(x.[No of Employees]) as Number, x.Salary as Salary
from
(
Select Count(Id) as [No of Employees], Salary
from tblPerson
Group by Salary
Having Salary = MAX(Salary)
)x
where x.[No of Employees]=3
Now I know this is a kind of long and complex way of doing it, but I was trying to do it using a derived table. But I am getting the error:
"Column 'x.Salary' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause"
My question is, why am I getting this particular error since the main query is a simple Select statement with a where clause. Isn't it??
Mainly, aggregate functions work only with other aggregate functions or grouped by columns.
Why? Because an aggregate function needs to know the set of values to do calculation with.
In you case, the max() will want to use all the data available for the calculation and display a single result (single row) and the other column will want to be displayed row by row. So there's a conflict.
Thank you all. Every answer helped me. However, I think I found a pretty simple way to do it:
Select top 1 Count(Id) as [No of Employees], salary
from tblPerson
Group by Salary
Order by [No of Employees] DESC
select count(*) from tblPerson where salary=(select max(salary) from tblPerson)
You get the error because 'max' is an aggregation, while you have nothing to aggregate the number by.
Select Max(x.[No of Employees]) as Number, x.Salary as Salary
from
(
Select Count(Id) as [No of Employees], Salary
from tblPerson
Group by Salary
Having Salary = MAX(Salary)
)x
---------
Group by Salary -- all other items in your select statement
---------
where x.[No of Employees]=3
however, you can also use a temporary table or variable to find the persons.
To solve this via a variable, you could do the following
declare #maxSalary Decimal
set #maxSalary = (Select max(salary) from tblperson) --insert the max value into a variable
Then either aggregate the persons (or do some other logic):
Select ID from tblperson where salary = #maxSalary
The reason for not using a group by is that using a variable is more efficient, as you search the table instead of aggregating over it.
Create a CTE (RESULT), and using DENSE_RANK function, get the highest salary, together with the EmployeeID's.
The first row of the RESULT table will give the highest salary.
Using the aggregate function COUNT, get the number of Employees with the highest Salary.
with RESULT (EmployeeID, Salary, DenseRank) as
(select EmployeeID, Salary,
DENSE_RANK() over (ORDER BY SALARY DESC) AS DenseRank
from Employee)
select TOP 1 Salary,
(select COUNT(EmployeeID)
from Employee
where Salary = (select TOP 1 Salary)
from RESULT
where DenseRank = 1)
)
from RESULT
where DenseRank = 1;
I have a table that is called: customers.
I'm trying to get the name and the salary of the people who have the maximum salary.
So I have tried this:
SELECT name, salary AS MaxSalary
FROM CUSTOMERS
GROUP BY salary
HAVING salary = max(salary)
Unfortunately, I got this error:
Column 'CUSTOMERS.name' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
I know I should add the name column to the group by clause, but I get all the records of the table.
I know that I can do it by:
SELECT name, salary
FROM CUSTOMERS
WHERE salary = (SELECT MAX(salary) FROM CUSTOMERS)
But I want to achieve it by group by and having clauses.
This requirement isn't really suited for a group by and having solution. The easiest way to do so, assuming you're using a modern-insh version of MS SQL Server, is to use the rank window function:
SELECT name, salary
FROM (SELECT name, salary, RANK() OVER (ORDER BY salary DESC) rk
FROM customers) c
WHERE rk = 1
Mureinik's answer is good with rank, but if you didn't want a windowed function for whatever reason, you can just use a CTE or a subquery.
with mxs as (
select
max(salary) max_salary
from
customers
)
select
name
,salary
from
customers cst
join mxs on mxs.max_salary = cst.salary
There was no need to use group by and having clause there, you know. But if you want to use them then query should be
SELECT name, salary
FROM CUSTOMERS
GROUP BY salary
having salary = (select max(salary) from CUSTOMERS)
To simplify, if I had a table called 'employee':
id INT NOT NULL PRIMARY KEY,
salary FLOAT,
department VARCHAR(255)
I want to perform and query where I retrieve the minimum salary in each department.
So my query is something like this:
SELECT employee.ID, MIN(employee.salary), employee.department
FROM employee
GROUP BY employee.department
But regardless of which records are found. The ID values in the result set are renamed to 1,2,3.... up to however many records (departments) exist in the result set.
How can I maintain the actual ID's of the employees after performing the AGGREGATE function and GROUP BY?
You can't. Think about it, If a Department has 20 employees, and for that department, there are three employees that have the same minimum salary, which EmployeeId do you want the the query output to display? if it was guaranteed that there was only one employee in each dept with that lowest salary, then it can be done by selecting the specific employee records where the salary is the minimum value for each Department:
Select EmployeeID
From Employee e
Where Salary =
(Select Min(Salary) From EMployee
Where DepartmentId = e.DepartmentId)
but this will return multiple records per department when more than one employee has that min salary level.
I would guess you're using MySQL or SQLite, because your query is ambiguous and isn't allowed by standard SQL or most brands of RDBMS. MySQL and SQLite are more permissive, so it's your responsibility to resolve the ambiguity.
Here's my usual fix:
SELECT e1.ID, e1.salary, e1.department
FROM employee e1
LEFT OUTER JOIN employee e2 ON (e1.department = e2.department
AND e1.salary > e2.salary)
WHERE e2.department IS NULL;
Here's another solution that gives the same result:
SELECT e1.ID, e1.salary, e1.department
FROM employee e1
JOIN (SELECT e2.department, MIN(e2.salary) AS min_salary
FROM employee e2 GROUP BY e2.department) d
ON (e1.salary = d.min_salary);
Both of these give multiple rows per department if there are multiple employees in the department with identical minimal salaries. You need to decide how to resolve that case, because it's not clear from your problem description.
Your script is invalid:
SELECT employee.ID, MIN(employee.salary), employee.department
FROM employee
GROUP BY employee.department
Instead, look at this:
SELECT MIN(employee.salary), employee.department
FROM employee
GROUP BY employee.department
If you need the employee id as well, then you need to use a subquery.
This will do the trick:
SELECT employee.department, MIN(employee.salary), employee.ID
FROM employee
GROUP BY 1
In modern SQL Server releases (and other reasonably powerful and modern SQL engines), SQL "Window functions" are probably the best alternatives (to be preferred over subqueries and self-joins) to do what you desire:
SELECT ID, salary, department
FROM employee
WHERE 1 = ROW_NUMBER() OVER(PARTITION BY department ORDER BY salary ASC)
This works when, if multiple employees have the same (department-minimal) salary, you want just a "random-ish" one of them (you can add criteria to the ORDER BY if you want one picked by some specific criteria); look into RANK, instead of ROW_NUMBER, if you want all.