Formatted question description: https://leetcode.ca/all/1280.html

1280. Students and Examinations

Level

Easy

Description

Table: Students

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| student_id    | int     |
| student_name  | varchar |
+---------------+---------+
student_id is the primary key for this table.
Each row of this table contains the ID and the name of one student in the school.

Table: Subjects

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| subject_name | varchar |
+--------------+---------+
subject_name is the primary key for this table.
Each row of this table contains the name of one subject in the school.

Table: Examinations

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| student_id   | int     |
| subject_name | varchar |
+--------------+---------+
There is no primary key for this table. It may contain duplicates.
Each student from the Students table takes every course from Subjects table.
Each row of this table indicates that a student with ID student_id attended the exam of subject_name.

Write an SQL query to find the number of times each student attended each exam.

Order the result table by student_id and subject_name.

The query result format is in the following example:

Students table:
+------------+--------------+
| student_id | student_name |
+------------+--------------+
| 1          | Alice        |
| 2          | Bob          |
| 13         | John         |
| 6          | Alex         |
+------------+--------------+
Subjects table:
+--------------+
| subject_name |
+--------------+
| Math         |
| Physics      |
| Programming  |
+--------------+
Examinations table:
+------------+--------------+
| student_id | subject_name |
+------------+--------------+
| 1          | Math         |
| 1          | Physics      |
| 1          | Programming  |
| 2          | Programming  |
| 1          | Physics      |
| 1          | Math         |
| 13         | Math         |
| 13         | Programming  |
| 13         | Physics      |
| 2          | Math         |
| 1          | Math         |
+------------+--------------+
Result table:
+------------+--------------+--------------+----------------+
| student_id | student_name | subject_name | attended_exams |
+------------+--------------+--------------+----------------+
| 1          | Alice        | Math         | 3              |
| 1          | Alice        | Physics      | 2              |
| 1          | Alice        | Programming  | 1              |
| 2          | Bob          | Math         | 1              |
| 2          | Bob          | Physics      | 0              |
| 2          | Bob          | Programming  | 1              |
| 6          | Alex         | Math         | 0              |
| 6          | Alex         | Physics      | 0              |
| 6          | Alex         | Programming  | 0              |
| 13         | John         | Math         | 1              |
| 13         | John         | Physics      | 1              |
| 13         | John         | Programming  | 1              |
+------------+--------------+--------------+----------------+
The result table should contain all students and all subjects.
Alice attended Math exam 3 times, Physics exam 2 times and Programming exam 1 time.
Bob attended Math exam 1 time, Programming exam 1 time and didn't attend the Physics exam.
Alex didn't attend any exam.
John attended Math exam 1 time, Physics exam 1 time and Programming exam 1 time.

Solution

If a student doesn’t attend any exam of a subject, then the result table should have an entry of this student and this subject with 0 attended exams.

Join tables Students and Subjects, and then use the joined table to join table Examinations using fields student_id and subject_name. After the tables are joined, select student_id, student_name, subject_id and count(Examinations.student_id) as attended_exams. Use group by and order by accordingly.

# Write your MySQL query statement below
select Students.student_id, student_name, Subjects.subject_name, count(Examinations.student_id) as attended_exams
    from (Students join Subjects on 1=1) left join Examinations
    on (Students.student_id, Subjects.subject_name) = (Examinations.student_id, Examinations.subject_name)
    group by Students.student_id, Students.student_name, Subjects.subject_name
    order by Students.student_id, Subjects.subject_name;

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