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I'm new to working with analytic functions.

 DEPT EMP SALARY ---- ----- ------ 10 MARY 100000 10 JOHN 200000 10 SCOTT 300000 20 BOB 100000 20 BETTY 200000 30 ALAN 100000 30 TOM 200000 30 JEFF 300000 

I want the department and employee with minimum salary.

Results should look like:

 DEPT EMP SALARY ---- ----- ------ 10 MARY 100000 20 BOB 100000 30 ALAN 100000 

EDIT: Here's the SQL I have (but of course, it doesn't work as it wants staff in the group by clause as well):

 SELECT dept, emp, MIN(salary) KEEP (DENSE_RANK FIRST ORDER BY salary) FROM mytable GROUP BY dept 
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4 Answers 4

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I think that the Rank() function is not the way to go with this, for two reasons.

Firstly, it is probably less efficient than a Min()-based method.

The reason for this is that the query has to maintain an ordered list of all salaries per department as it scans the data, and the rank will then be assigned later by re-reading this list. Obviously in the absence of indexes that can be leveraged for this, you cannot assign a rank until the last data item has been read, and maintenance of the list is expensive.

So the performance of the Rank() function is dependent on the total number of elements to be scanned, and if the number is sufficient that the sort spills to disk then performance will collapse.

This is probably more efficient:

select dept, emp, salary from ( SELECT dept, emp, salary, Min(salary) Over (Partition By dept) min_salary FROM mytable ) where salary = min_salary / 

This method only requires that the query maintain a single value per department of the minimum value encountered so far. If a new minimum is encountered then the existing value is modified, otherwise the new value is discarded. The total number of elements that have to be held in memory is related to the number of departments, not the number of rows scanned.

It could be that Oracle has a code path to recognise that the Rank does not really need to be computed in this case, but I wouldn't bet on it.

The second reason for disliking Rank() is that it just answers the wrong question. The question is not "Which records have the salary that is the first ranking when the salaries per department are ascending ordered", it is "Which records have the salary that is the minimum per department". That makes a big difference to me, at least.

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1 Comment

Thank you David. After considering its benefits, I refactored to your solution.
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I think you were pretty close with your original query. The following would run and do match your test case:

SELECT dept, MIN(emp) KEEP(DENSE_RANK FIRST ORDER BY salary, ROWID) AS emp, MIN(salary) KEEP (DENSE_RANK FIRST ORDER BY salary, ROWID) AS salary FROM mytable GROUP BY dept 

In contrast to the RANK() solutions, this one guarantees at most one row per department. But that hints at a problem: what happens in a department where there are two employees on the lowest salary? The RANK() solutions will return both employees -- more than one row for the department. This answer will pick one arbitrarily and make sure there's only one for the department.

2 Comments

Yeah, that's a good point on the multiple records. The Min() method(s) will retrieve all the duplicates ... be trickier to get a single record back for those if one were needed.
Excellent elaboration -- especially if analysis being posed is more concerned with the value of the minimum. If identifying attributes of the minimum were needed, preserving duplicates would seem to be desirable.
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You can use the RANK() syntax. For example, this query will tell you where an employee ranks within their department with regard to how large their salary is:

SELECT dept, emp, salary, (RANK() OVER (PARTITION BY dept ORDER BY salary)) salary_rank_within_dept FROM EMPLOYEES 

You could then query from this where salary_rank_within_dept = 1:

SELECT * FROM ( SELECT dept, emp, salary, (RANK() OVER (PARTITION BY dept ORDER BY salary)) salary_rank_within_dept FROM EMPLOYEES ) WHERE salary_rank_within_dept = 1 

15 Comments

Perfect! I didn't know about RANK() yet. Thanks.
I didn't even know about RANK() until yesterday! :)
I'm downvoting this for the reasons I outlined in my own answer: I think it's probably inefficient, and I think that the query is not a good match to the exact question being asked. I'm not saying that it won't give the correct answer, just that it doesn't express the logic of the question very well.
@David: Pretty obvious from the answer timestamps that you lifted my answer while voting me down.
Dudes, if you want to have a flame war, get a blog. I thank everyone for helping me with my query. After reviewing David's solution, I've refactored to it and marked it as the answer because I believe it to be a better answer. I now urge everyone to move on.
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select e2.dept, e2.emp, e2.salary from employee e2 where e2.salary = (select min(e1.salary) from employee e1) 

1 Comment

That will give you one record - the minimum for the entire table. You need to group by the department in your subselect.

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