As I was learning SQL statements I encountered one example (regarding the demo SCOTT database), I have no idea how to solve.
In which department(s) are all salgrades present?
My most promising approach is to group all salgrades and departments in the joined tables emp , dept and salgrade :
SELECT s.grade AS "Salgrade",
d.dname AS "Department ID"
FROM emp e INNER JOIN dept d ON(e.deptno = d.deptno)
INNER JOIN salgrade s ON(e.sal BETWEEN s.losal AND s.hisal)
GROUP BY d.dname, s.grade
Executing this gives me the following results:
If I could group this another time by department, COUNT(*) could give me the number of different salgrades per department . Then I could compare this number (with HAVING ) to the following subselect:
(SELECT COUNT(*)
FROM salgrade)
Is there any possibility to group a table which already contains GROUP BY ? Is there another (better) approach I could use?
I am using an apex-oracle-server with "Application Express 4.2.4.00.07"
Minor change from your version, by removing the grouping inside, and this version, first generates, salgrade
and department
of all employees, and then doing a grouping outside, counting distinct salary grades.
SELECT Department_ID
FROM
(
SELECT s.grade AS Salgrade,
d.dname AS Department_ID
FROM emp e
INNER JOIN dept d ON(e.deptno = d.deptno)
INNER JOIN salgrade s ON(e.sal BETWEEN s.losal AND s.hisal)
)
GROUP BY Department_ID
HAVING COUNT(distinct Salgrade) = ( SELECT count(1) FROM salgrade);
I found an even easier solution now:
SELECT d.dname
FROM emp e INNER JOIN dept d ON(e.deptno = d.deptno)
INNER JOIN salgrade s ON(e.sal BETWEEN s.losal AND s.hisal)
GROUP BY d.dname
HAVING COUNT(DISTINCT s.grade) = (SELECT COUNT(*) FROM salgrade);
Simple way would be - if performance is not a problem.
SELECT
COUNT(DISTINCT [Salgrade]) AS [COUNT]
,[Department ID]
FROM (SELECT s.grade AS "Salgrade",
d.dname AS "Department ID"
FROM emp e INNER JOIN dept d ON(e.deptno = d.deptno)
INNER JOIN salgrade s ON(e.sal BETWEEN s.losal AND s.hisal)
GROUP BY d.dname, s.grade) DEPT_SALE
GROUP BY [Department ID]
There could be better solutions though if we know more of your base tables - emp
& salegrade
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