Question

I am working on my final project and this is the query that returns the list of all id’s, names and days.

select 
 c.id as id, i.full_name as n, day as d
 from
 (instructors i
 inner join courses c ON c.instructor_id = i.id)
 inner join exam_schedules e ON c.id = e.course_id

results samlpe:

id  name    day
9977    SCOTT, RAYMOND C.   wednesday
9467    FERNANDEZ, CYNTHIA E.   wednesday
9468    FERNANDEZ, CYNTHIA E.   wednesday
9493    FERNANDEZ, CYNTHIA E.   wednesday
9538    STEPHENS, RONNIE T. wednesday
9539    MYERS, RONNIE Y.    wednesday
9554    GARRISON, EUGENE A. wednesday

How can I get the list of all ID’s and count of combinations of name+day ?

Example above should produce:

9977    1
9467    3
9468    3
9493    3
9538    1
9539    1
9554    1

Answer 1

I used temporary table for clarity. It is possible to merge this into one query, but it is very hard to do without the real schema:

CREATE TABLE tmp2 ( id int, name_day varchar(80) );

INSERT INTO tmp2 (id, name_day) (SELECT c.id AS id, CONCAT(i.full_name, i.day) AS name_day
     FROM (instructors i
         INNER JOIN courses c ON c.instructor_id = i.id)
         INNER JOIN exam_schedules e ON c.id = e.course_id
         );

SELECT id, counter FROM 
    (SELECT name_day, count(*) AS counter FROM tmp2 GROUP BY name_day) AS t1
    INNER JOIN
    tmp2
    ON t1.name_day = tmp2.name_day;

DROP TABLE tmp2;