Interview Question:
Write a single SELECT statement that returns the number of employees who get no commission broken down by department. (Use scott.emp table)
Level:
Intermediate
Expected Result:
| DEPTNO | NO_COMM_COUNT |
|---|---|
| 10 | 3 |
| 20 | 5 |
| 30 | 3 |
Solutions
A very typical attempt to solve this problem results in the following query:
SELECT deptno, COUNT(*) no_comm_count FROM scott.emp WHERE comm IS NULL OR comm=0 GROUP BY deptno ORDER BY 1
Yes, the result looks correct, but is the query correct?
The answer is NO! It would become apparent if we had a department where all employees get paid commission, so the number of those who does not would be 0.
Let’s change the requirement a bit – we will show all department and number of employees hired on Friday:
SELECT deptno, COUNT(*) fri_count FROM scott.emp WHERE TO_CHAR(hiredate, 'DY')='FRI' GROUP BY deptno ORDER BY 1
The result of this query is clearly not what we want:
| DEPTNO | FRI_COUNT |
|---|---|
| 30 | 2 |
We would expect the following instead:
| DEPTNO | FRI_COUNT |
|---|---|
| 10 | 0 |
| 20 | 0 |
| 30 | 2 |
Why don’t we get the departments 10 and 20? The answer is very simple – because we filter “all” those department rows with our WHERE clause. So how should we work around?
Let’s start with more intuitive but less efficient approaches – we will use the same query as before and UNION it with another query that returns “empty” departments. Essentially, the original problem transforms into a new one – find all department where no employees were hired on Friday.
Strategy #1: Using UNION ALL with multi-column non-correlated subquery:
SELECT deptno, COUNT(*) fri_count
FROM scott.emp
WHERE TO_CHAR(hiredate, 'DY')='FRI'
GROUP BY deptno
UNION ALL
SELECT deptno, 0 fri_count
FROM scott.emp
WHERE (deptno, 'FRI') NOT IN (SELECT deptno, TO_CHAR(hiredate, 'DY')
FROM scott.emp)
GROUP BY deptno
ORDER BY 1
| DEPTNO | FRI_COUNT |
|---|---|
| 10 | 0 |
| 20 | 0 |
| 30 | 2 |
Strategy #2: Using UNION ALL with ALL predicate on correlated subquery:
SELECT deptno, COUNT(*) fri_count
FROM scott.emp
WHERE TO_CHAR(hiredate, 'DY')='FRI'
GROUP BY deptno
UNION ALL
SELECT deptno, 0 no_comm_count
FROM scott.emp a
WHERE 'FRI'!=ALL(SELECT TO_CHAR(hiredate, 'DY')
FROM scott.emp b
WHERE a.deptno=b.deptno)
GROUP BY deptno
ORDER BY 1
It is apparent that the ALL predicate ensures that no employees were hired on Friday.
Now we will mimic the behavior of the UNION ALL operator using LEFT JOIN:
Strategy #3: Using LEFT JOIN:
SELECT a.deptno, COUNT(DISTINCT b.empno) fri_count
FROM scott.emp a LEFT JOIN scott.emp b ON a.deptno=b.deptno
AND TO_CHAR(b.hiredate, 'DY')='FRI'
GROUP BY a.deptno
ORDER BY 1
COUNT(DISTINCT …) is needed to handle a Cartesian Product as the join by deptno column produces many to many relationship, i.e. Cartesian product.
Strategy #4: Generic substitution technique for an outer-join using UNION ALL
WITH e AS ( SELECT deptno, COUNT(*) fri_count FROM scott.emp WHERE TO_CHAR(hiredate, 'DY') = 'FRI' GROUP BY deptno UNION ALL SELECT deptno, 0 FROM scott.emp GROUP BY deptno ) SELECT deptno, MAX(fri_count) fri_count FROM e GROUP BY deptno ORDER BY 1
All the above techniques may look cool but they are clearly an overkill for such a simple problem. There is a simple rule worth remembering:
If you need to conditionally aggregate all records in the table but you fail doing so due to a WHERE clause filter, consider moving the filter into the GROUP function you use in SELECT.
Strategy #5: Conditional Aggregation
SELECT deptno, COUNT(DECODE(TO_CHAR(hiredate, 'DY'), 'FRI', 1)) fri_count FROM scott.emp GROUP BY deptno ORDER BY 1
Alternatively, you can use CASE function inside of COUNT. It is especially convenient for our original question/problem, i.e. to count employees who is not paid a commission:
SELECT deptno, COUNT(CASE WHEN LNNVL(comm>0) THEN 1 END) no_comm_count FROM scott.emp GROUP BY deptno ORDER BY 1
| DEPTNO | NO_COMM_COUNT |
|---|---|
| 10 | 3 |
| 20 | 5 |
| 30 | 3 |
This approach is the most efficient as it makes Oracle scanning the emp table only once.
Notice the use of the LNNVL function. You can read more about it in my recent post here.
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Suggested Reading:
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