Access SQL Can't Create Two Grouped Averages - sql

Apologies for my simple problem, I am an absolute novice. I have the following code in separate queries
I am attempting to display 3 columns, the average male salary for a set job, average female salary for a set job and the JobID. Separately these queries work however I cannot work out how to combine them.
I have tried multiple solutions from this site for example trying to put multiple select statements inside
and also by using a 'union' solution however cannot get either to work.union This simply compiles them into a single column and sorts via salary not JobID.
SELECT Round(Avg(Salary)) AS AverageMaleSalary, JobID
FROM Employee WHERE Gender = "M"
GROUP BY JobID;
SELECT Round(Avg(Salary)) AS AverageFemaleSalary, JobID
FROM Employee WHERE Gender = "F"
GROUP BY JobID;

You could use conditional aggregation
SELECT JobId,ROUND(AVG(IIF(Gender='F', Salary, NULL))) AS AverageFemaleSalary
,ROUND(AVG(IIF(Gender='M', Salary, NULL))) AS AverageMaleSalary
FROM Employee
GROUP BY JobId;

Related

Using ROUND, AVG and COUNT in the same SQL query

I need to write a query where I need to first count the people working in a department, then calculate the average people working in a department and finally round it to only one decimal place. I tried so many different variations.
That's what I got so far although it's not the first one I tried but I always get the same error message. (ORA-00979 - not a group by expression)
SELECT department_id,
ROUND(AVG(c.cnumber),1)
FROM employees c
WHERE c.cnumber =
(SELECT COUNT(c.employee_id)
FROM employees c)
GROUP BY department_id;
I really don't know what do to at this point and would appreciate any help.
Employees Table:
Try this (Oracle syntax) example from your description:
with department_count as (
SELECT department_id, COUNT(c.employee_id) as employee_count
FROM employees c
group by department_id
)
SELECT department_id,
ROUND(AVG(c.employee_count),1)
FROM department_count c
GROUP BY department_id;
But this query not make sense. Count is integer, and count return one number for one department in this case AVG return the same value as count.
Maybe you have calculate number of employee and averange of salary on department?

SQL query from Lynda.com

I have a question about two queries. Will these two queries give the same result? I am trying to find the average salary by department:
Select s1.department, avg(s1.salary)
From
(Select department, salary
From staff
Where salary > 100000) s1
Group by s1.department
vs
select department, avg(salary) as avg_salary
from staff
where salary > 100000
group by department
Yes, it gives the same amounts back.
the bottom query gets data from a sub select which gets its data from the table, whereas the top query gets it straight from the table itself.
There are no additional filters in there. So the result will be the same.
you can test it out however, don't take my word for it.

SQL: A Count that merges/joins two columns

A company wants to know what has happened since the beginning of 1999 on hiring. The human resources manager has asked you to
produce a count of the employees hired since then, broken down by both age and gender simultaneously (i.e. 17 males, 25 females, etc.). Write a query that does that.
This is what I have so far. I couldn't figure out how to merge two columns simultaneously. Any thoughts?
SELECT EmployeeID, COUNT(*) AS "Number of employees"
FROM Employee
WHERE Age and Gender
GROUP BY EmployeeID
HAVING COUNT(*) BEGIN = 1999
I am not pretty much sure about your requirement but the right way to write a query is as following and this may fulfill your requirement with slight changes...
SELECT Gender, COUNT(*) AS "Number of employees", Age
FROM Employee
WHERE year(column_name) >= 1999 --Give a real date column here
GROUP BY Age, Gender
Please provide table structure and complete desired output for the exact answer.

Can I write a condition on a single column name of a relation in HAVING clause

Considering the following relational schema
customers(id, name, age, address, salary)
I tried a query
SELECT SUM(salary), age FROM customers
GROUP BY age HAVING age > 23 ; ...(1)
I was surprised to see that it worked fine and that I could write a single column condition also in HAVING clause.
Even this is also working
SELECT SUM(salary), age FROM customers
GROUP BY age, salary HAVING age > 23 AND salary >2000; ...(2)
Otherwise, I should have written it like this : (using WHERE clause)
SELECT SUM(salary), age FROM customers
WHERE age > 23 GROUP BY age; ...(3)
And
SELECT SUM(salary), age FROM customers
WHERE age > 23 AND salary >2000 GROUP BY age, salary ; ..(4)
But when I tried with more combinations I found that
that column name must be present in GROUP BY clause also on which condition is applied in HAVING clause.
Am I correct or is it possible to write a single column condition in HAVING clause in any other way also ?
Why is it working because I had earlier studied that we can write only conditions on Aggregate functions in HAVING clause.
You're generally correct. Important thing is to understand grouping at all.
When using GROUP BY, server scans 'rows' and buckets them into some 'groups'. Then every 'group' works as a single new row. When operating these 'new lines' - in SELECT, HAVING or ORDER clauses - server needs to know 'attribute values' of them. These attribute values are aggregations of rows' attribute values or expressions with these aggregations.
When some attribute or expression used in the GROUP BY clause, it's aggregation values are quite deterministic, so server give us ability to simplify process. We can write something like
SELECT object_type, count(*)
FROM user_objects
GROUP BY object_type
HAVING MAX(object_type) like '%O%'
ORDER BY MIN(object_type)
It would work fine if we do this. But we can write simply
SELECT object_type, count(*)
FROM user_objects
GROUP BY object_type
HAVING object_type like '%O%'
ORDER BY object_type
which means exact the same. If column does not mentioned into the GROUP BY values - rule above became not true, so we cannot use it directly, without aggregation.
SELECT SUM(salary), age
FROM customers
GROUP BY age, salary
HAVING age > 23 AND salary >2000;
This gives you one record per age and salary, as you group by these. Later you remove some of the result lines. The sum of the salary is of course the salary itself.
If these are your records for instance:
salary age something
1000 30 100
1000 30 200
2000 30 300
2000 40 400
then you group like this:
salary age something
1000 30 100
200
2000 30 300
2000 40 400
For the group 1000/30 the sum(something) is 300 and avg(something) is 150. But the sum(salary) is 1000 and avg(salary) is 1000 and min(salary) is 1000, and so on, because it is just one salary value you are talking about.
The HAVING clause then removes lines from the result where age is over 23 and salary over 2000. You could have removed these records from evaluation by using a WHERE clause instead, thus saving the dbms some work. But you made the dbms collect all age and salary groups first, only to say which ones you dismiss afterwards.
I agree though that it would be better the DBMS raised an error telling you that sum(salary) makes no sense as it is just the one salary of the group.

Using aggregate function for seprate rows

There is a table with name jobs where for different department number salary is giving. Same department can have more than one job so the salary may vary. Now i want to solve this query:
"Find the average salaries for each department without displaying the respective department numbers."
Here i should use avg but how to use it so that i can get my result of each department no separately?
I think what you're looking for is the GROUP BY clause. If I understand your question correctly then something like this should do the trick.
SELECT AVG(salary) FROM table_name GROUP BY department