SQL如何为每个字段计数表

Question

I have 3 tables which is:

Courses

• courses_id
• name

QnAs

• qna_id
• student_id
• courses_id
• name
• question

Students

• student_id
• name

Now I'm trying to count how many qna's there are for each courses. How do i make the query?

I've tried doing this :

SELECT (SELECT COUNT(qna_id) AS Expr1
              FROM    QnAs) AS Count
FROM   QnAs AS QnAs_1 CROSS JOIN
             Courses
GROUP BY Courses.courses_id

It does counts how many QnA's there are but not for each Courses

The output i got is each Courses names and QnAs count number but what i want is the QnA's number for each of the Courses

Answer 1

Why not just use GROUP BY ?

SELECT q.courses_id, COUNT(qna_id) as cnt
FROM QnAs q
GROUP BY q.courses_id;

Answer 2

It seems you merely want to aggregate QNAs by course ID:

select courses_id, count(*)
from qnas
group by courses_id
order by courses_id;

Along with the course names:

select c.course_id, c.name, coalesce(q.cnt, 0) as qna_count
from courses c
left join
(
  select courses_id, count(*) as cnt
  from qnas
  group by courses_id
) q on q.course_id = c.course_id
order by c.course_id;

Answer 3

This is not an answer, but just an explanation what your query does.

In your own query you first cross join all QnAs with all courses for no apparent reason, thus getting all possible combinations. So with two courses, each with three QNAs (that makes six QNAs in total), you'd construct 2 x 6 = 12 rows.

For each of these rows you select the total number of rows in the QNA table, which is six in above example. So you'd select 12 rows, all showing the number 6.

But then you group by course ID, thus ending up with two rows only in my example. You should apply an aggregate function on your subquery, eg MAX or SUM , but you don't, which makes your query invalid (because you are dealing with many rows, but treat this as if it were a single value). MySQL however silently applies ANY_VALUE , so your query becomes:

SELECT
  ANY_VALUE( (SELECT COUNT(*) FROM QnAs) ) AS Count
FROM QnAs AS QnAs_1 
CROSS JOIN Courses
GROUP BY Courses.courses_id;

I hope this explanation helps you understand how joins and aggregation work. You may want to set ONLY_FULL_GROUP_BY mode ( https://dev.mysql.com/doc/... ) in order to have MySQL report the syntax error instead of silently "fixing" the query by applying ANY_VALUE .