i have a table that "users"
i want to count their salary who have ID 1 and 2
how i can count this in mysql
Have a look at Aggregate Functions in MySQL and GROUP BY (Aggregate) Functions
to get total salary of both
SELECT SUM(salary) FROM users WHERE id IN(1, 2);
and to get individual sum of salary
SELECT SUM(salary) FROM users WHERE id IN(1, 2) group by id;
SELECT SUM(salary) FROM users WHERE id IN(1, 2);
I shall refer you to counting rows from the mysql documentation
probably goes like:
SELECT COUNT(*) FROM users WHERE id IN (1,2);
although if you're actually trying to sum the salaries then you should refer to group by functions
SELECT id, SUM(salary) FROM users WHERE id IN (1,2) GROUP BY id;
Related
Well, that is, there are two Johns and one Quill, you need to output the number of those people who have the same name. In one column there should be the total number of students with the same names
SELECT COUNT(id) as count
FROM student
GROUP BY LOWER(first_name) HAVING COUNT(LOWER(first_name)) > 1;
it will output for each name the count, how to make the total?
In order to get the total, select from your query result and add the counts up.
SELECT SUM(cnt)
FROM
(
SELECT COUNT(*) AS cnt
FROM student
GROUP BY LOWER(first_name)
HAVING COUNT(*) > 1
) counts;
Please, try the following:
SELECT SUM(COUNT(id)) as Total
FROM student
GROUP BY LOWER(first_name) HAVING COUNT(LOWER(first_name)) > 1;
I have employee table with emp id (emp_id) and department (dep_id) fields. An employee could be working in more than one Department. I want to write a sql query to display unique emp_ids who work in more than one department.
Pl help me to write sql query.
Thx
Answered here: SQL query for finding records where count > 1
You need to use count, group by and having like this.
select emp_id, count(dep_id)
from employee_department
group by emp_id
having count(dep_id)>1
Query
SELECT COUNT(*)
FROM
(
SELECT id_employee, COUNT(*) AS CNT
FROM Department_Employee
GROUP BY id_employee
) AS T
WHERE CNT > 1
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)
I've retrieved a count of the duplicates and their occurrences using the below code
select empID, count(empID) AS DUPLICATEempID
from employees
group by empID
having count (empID) > 1
I now want the table to include the number of rows returned (i.e. insert the number on the table returned)
thanks in advance.
In SAS, you can do this with a subquery:
select empId, DUPLICateempID, count(*) as NumDuplicates
from (select empID, count(empID) AS DUPLICATEempID
from employees
group by empID
having count (empID) > 1
) t
When you have an aggregation function without a group by, it applies the function to the whole table and re-merges the results.