query using inner joins and having - mysql

Question

These are my tables:
mysql> select * from professor;
+-------+--------+--------+--------+------+
| empid | name | status | salary | age |
+-------+--------+--------+--------+------+
| 1 | Arun | 1 | 2000 | 23 |
| 2 | Benoy | 0 | 3000 | 25 |
| 3 | Chacko | 1 | 1000 | 36 |
| 4 | Divin | 0 | 5000 | 32 |
| 5 | Edwin | 1 | 2500 | 55 |
| 7 | George | 0 | 1500 | 46 |
+-------+--------+--------+--------+------+
6 rows in set (0.00 sec)
mysql> select * from works;
+----------+-------+---------+
| courseid | empid | classid |
+----------+-------+---------+
| 1 | 1 | 10 |
| 2 | 2 | 9 |
| 3 | 3 | 8 |
| 4 | 4 | 10 |
| 5 | 5 | 9 |
| 6 | 1 | 9 |
| 2 | 3 | 10 |
| 2 | 1 | 7 |
| 4 | 2 | 6 |
| 2 | 4 | 6 |
| 2 | 5 | 2 |
| 7 | 5 | 6 |
| 3 | 5 | 2 |
| 6 | 4 | 10 |
| 2 | 7 | 1 |
+----------+-------+---------+
15 rows in set (0.00 sec)
mysql> select * from course;
+----------+------------+--------+
| courseid | coursename | points |
+----------+------------+--------+
| 1 | Maths | 5 |
| 2 | Science | 1 |
| 3 | English | 6 |
| 4 | Social | 4 |
| 5 | Malayalam | 20 |
| 6 | Arts | 25 |
| 7 | Biology | 20 |
+----------+------------+--------+
7 rows in set (0.00 sec)
The question is :
Return those courses that have been taught by all professors.
The query I tries is:
select course.coursename from
course inner join works
on course.courseid=works.empid
group by works.courseid
having works.empid in (select empid from professor);
I am getting an error like this:
Unknown column 'works.empid' in 'IN/ALL/ANY subquery'
Pls help me out with the query.
http://sqlfiddle.com/#!2/4b197/5

Answer 1

Apart from the reasons why your current query gives you an error, which #GordonLinoff explained in detail, one way to achieve the desired result
Return those courses that have been taught by all professors.
is
SELECT c.*
FROM
(
SELECT courseid
FROM works
GROUP BY courseid
HAVING COUNT(DISTINCT empid) =
(
SELECT COUNT(*)
FROM professor
)
) q JOIN course c
ON q.courseid = c.courseid
Note: Thanks to #eggyal it's worth to mention that this query operates on the assumption that referential integrity is intact meaning works table doesn't have orphaned records (rows where empid refers to a non-existent row in professor table) and technically returns courses taught by the same number of professors as currently exist in the professor table which in the case of intact referential integrity happen to be the courses we're looking for.
Output:
| COURSEID | COURSENAME | POINTS |
|----------|------------|--------|
| 2 | Science | 1 |
Here is SQLFiddle demo

Answer 2

First, the query is way off from what you want. I'm only going to address the error.
The error is very interesting. In short, MySQL allows something called hidden columns (which are described below). However, these only work in the having clause when they are included in the select clause. I hadn't known that.
The following two queries parse correctly (I'm skipping the middle part for brevity):
select course.coursename, works.empid
. . .
having works.empid = 1;
select course.coursename, works.empid
. . .
having works.empid = 1;
And yet, the following two fail with the same error:
select course.coursename
. . .
having works.empid = 1;
select course.coursename
. . .
having works.empid = 1;
The only difference s that the column is not mentioned in the select clause.
What is happening is that you are using a MySQL extension to the group by clause, sometimes called "Hidden Columns". You have columns in the select or having clause that are neither aggregation keys (course.courseid) nor surrounded by an aggregation function (say min(works.empid) or group_concat(works.empid). Apparently, MySQL only recognizes these columns in the having clause when they are already in the `select clause. At your stage of learning SQL, you just shouldn't do this. Following the documentation to turn off this extension and go to ANSI standard behavior:
To disable the MySQL GROUP BY extension, enable the ONLY_FULL_GROUP_BY
SQL mode. This enables standard SQL behavior: Columns not named in the
GROUP BY clause cannot be used in the select list or HAVING clause
unless enclosed in an aggregate function.
The way to fix the syntax problem is to use an aggregation function, something like:
select course.coursename
. . .
having min(works.empid) = 1;
select course.coursename
. . .
having min(works.empid) = 1;
This will get you no closer to having a working query, because yours is far from solving the problem. But it will fix the syntactic error.

Answer 3

Make a cross-join between course and professor to obtain every combination of courses and professors;
Make an outer join between that and works to identify which of those (course, professor) combinations actually exist;
Group by course and sum the number of such combinations that do not exist;
Filter the groups for only those where there are no such non-existent combinations.
Therefore:
SELECT course.*
FROM (course, professor) LEFT JOIN works USING (courseid, empid)
GROUP BY courseid
HAVING SUM(works.empid IS NULL) = 0

Answer 4

try this:
select c.coursename, w.empid from
course as c inner join works as w
on (c.courseid=w.empid)
group by w.courseid
having w.empid in (select p.empid from professor as p);

Answer 5

select course.coursename,works.courseid,works.empid from
course inner join works
on course.courseid=works.empid
group by works.courseid
having works.empid in (select empid from professor);
Its Working