MySql query - counting rows and percentages

Question

I'm trying to make a list of students and their efficiency with tests. In MySql database I have tables

users - table with students

id | name
_________
1  | Joe
2  | Marry
3  | Max
4  | Anna
----------

courses - table with courses

id | name
_____________
1  | Course 1
2  | Course 2
----------

questions - table with questions for each course. Row cours_id indicates in which course that question belongs to

id | cours_id | question
_________________________________
1  | 1        | Course 1 - question 1
2  | 1        | Course 1 - question 2
3  | 1        | Course 1 - question 3
4  | 1        | Course 1 - question 4
5  | 2        | Course 2 - question 1
6  | 2        | Course 2 - question 2
7  | 2        | Course 2 - question 3
8  | 2        | Course 2 - question 4

---

cours_invitations - every student gets invitation for a course. Row user_id shows the id of user who is invited to do a course. cours_id represents the id of the course which a student should do. When row status has value 0 that means that a student didn't started with the course (pending), and if it has value 1 that means that the student has started it (or done).

id | user_id | cours_id | status
________________________________
1  | 1       | 1        | 1
2  | 1       | 2        | 0
3  | 2       | 1        | 0
4  | 3       | 1        | 1
5  | 4       | 1        | 1
6  | 4       | 2        | 1

Example: Joe and Anna are invited to do Course 1 and Course 2, Marry and Max are invited to do only Course 1. Joe did Course 1 but not Course 2, Marry didn't do anything and Max did Course 1

courses_stats - are the statistics of the courses' questions which students have done. Status represent the accuracy of the answer. 0 stands for wrong answer and 1 for correct.

id | user_id | question_id | status

___________________________________
1  | 1       | 1           | 1
2  | 1       | 2           | 1
3  | 1       | 3           | 0
4  | 2       | 1           | 1
5  | 2       | 2           | 1
6  | 2       | 3           | 1
7  | 2       | 4           | 1
8  | 4       | 1           | 1
9  | 4       | 2           | 1
10 | 4       | 3           | 0
11 | 4       | 4           | 0
12 | 4       | 5           | 1
13 | 4       | 6           | 1

Example: Joe did 3 questions from the first course.Notice that he didn't do all the questions from that course and that one is incorrect.

Max did all the questions correct and Anna did all question from first course (half are incorrect) and half from second course (all correct)

I need a query with sudents names, perecentage of finished courses, percentage of correct answers of those courses which they did (not all courses) and possibility to order students by those percentages. Something like this:

name  | completed courses | completed questions
______________________________________________
Max   |100%               |100%
Anna  |100%               |50%
Joe   |50%                |50%
Marry |0%                 |0%

Is something like this even possible? Do I need more rows in tables for this query?

Answer 1

I think this should be what you need:

SELECT
    users.name,
    CONCAT(COUNT(
        DISTINCT CASE
        WHEN cours_invitations.status = 1 THEN
            cours_invitations.id
        ELSE
            NULL
        END
    ) / COUNT(
        DISTINCT cours_invitations.id
    ) * 100, '%') AS completed_courses,
    CONCAT(COUNT(
        DISTINCT CASE
        WHEN courses_stats.status = 1 THEN
            courses_stats.id
        ELSE
            NULL
        END
    ) / COUNT(DISTINCT questions.id) * 100, '%') AS completed_questions
FROM
    users
LEFT JOIN cours_invitations ON cours_invitations.user_id = users.id
LEFT JOIN questions ON cours_invitations.cours_id = questions.cours_id
AND cours_invitations.status = 1
LEFT JOIN courses_stats ON users.id = courses_stats.user_id
GROUP BY
    users.id
ORDER BY
    completed_courses DESC,
    completed_questions DESC

As a question back to you, why are the table names called cours_* rather than course_*?

Answer 2

Here you can find with table schemas and sample data, and result of query. MichaelRushton has perfect answer but course_stats has to be LEFT join I think. Because if student has cours_invitations but no course_stats, that query will not return that user.

http://sqlfiddle.com/#!2/019dc/1

SELECT
    users.name,
    COUNT(
        DISTINCT CASE
        WHEN course_invitations.status = 1 THEN
            course_invitations.id
        ELSE
            NULL
        END
    ) / COUNT(
        DISTINCT course_invitations.id
    ) * 100 AS completed_courses,
    COUNT(
        DISTINCT CASE
        WHEN courses_stats.status = 1 THEN
            courses_stats.id
        ELSE
            NULL
        END
    ) / COUNT(DISTINCT questions.id) * 100 AS completed_questions
FROM users
INNER JOIN course_invitations ON course_invitations.user_id = users.id
INNER JOIN questions ON course_invitations.cours_id = questions.cours_id
LEFT JOIN courses_stats ON users.id = courses_stats.user_id
GROUP BY
    users.id

Result:

NAME    COMPLETED_COURSES   COMPLETED_QUESTIONS
Joe 50  25
Marry   0   100
Max 100 0
Anna    100 50

Answer 3

I guess you meant "percentage of correct answers". OK, with clever using of count(distinct if(..)) construct you can avoid complex subqueries with different group-by clauses. For example, this code

count(distinct if(cours_invitations.status and courses_stats.status, 
                      NULL, questions.id)) 

counts the number of (distinct) questions where the condition cours_invitations.status and courses_stats.status is met. Using this trick, the whole query is as simple and elegant as this:

select users.name, 
    count(distinct if(cours_invitations.status, 
                         NULL, 
                         cours_invitations.cours_id)) 
        / count(distinct cours_invitations.cours_id) 
        * 100 as courses_completed,
    count(distinct if(cours_invitations.status and courses_stats.status, 
                          NULL, 
                          questions.id)) 
        / count(distinct if(cours_invitations.status, 
                               NULL, 
                               questions.id)) 
        * 100 as correct_answers
from users
    left join cours_invitations on users.id = cours_invitations.user_id
    left join questions using (cours_id)
    left join courses_stats on users.id = courses_stats.user_id 
                               and questions.id = courses_stats.question_id
group by users.id
order by correct_answers

I recommend to add the percentage sign outside mysql as it is much more elegant there, it would add uneccessary complexity to mysql query.