PostgreSQL - order by an array

Question

I have 2 tables - course that contains id and name of the courses and tagCourse that contains tags for each course.

course                    tagcourse
-----         


        

Answer 1

CREATE OR REPLACE FUNCTION search_by_tags(tags varchar[])
  RETURNS TABLE (id_course integer, name text, tag_ct integer) AS
$func$
   SELECT id_course, c.name, ct.tag_ct
   FROM  (
      SELECT tc.id_course, count(*)::int AS tag_ct
      FROM   unnest($1) x(tag)
      JOIN   tagcourse tc USING (tag)
      GROUP  BY 1                      -- first aggregate ..
      ) AS ct
   JOIN   course c USING (id_course)   -- .. then join
   ORDER  BY ct.tag_ct DESC            --  more columns to break ties?
$func$  LANGUAGE sql;

• Use unnest() to produce a table from your input array, like already demonstrated by @Clodoaldo.

• You don't need plpgsql for this. Simpler with a plain SQL function.

• I use unnest($1) (with positional parameter) instead of unnest(tags), since the later is only valid for PostgreSQL 9.2+ in SQL functions (unlike plpgsql). I quote the manual here:

In the older numeric approach, arguments are referenced using the syntax $n: $1 refers to the first input argument, $2 to the second, and so on. This will work whether or not the particular argument was declared with a name.

• count() returns bigint. You need to cast it to int to match the declared return type or declare the the returned column as bigint to begin with.

• Perfect occasion to simplify the syntax a bit with USING (equi-joins): USING (tag) instead of ON tc.tag = c.tag.

• It's regularly faster to first aggregate, then join to another table. Reduces the needed join operations.
  As per question of @Clodoaldo in the comments, here is an SQL Fiddle to demonstrate the difference.

• OTOH, if you aggregate after the join, you don't need a subquery. Shorter, but probably slower:

SELECT c.id_course, c.name, count(*)::int AS tag_ct
FROM   unnest($1) x(tag)
JOIN   tagcourse tc USING (tag)
JOIN   course     c USING (id_course)
GROUP  BY 1
ORDER  BY 3 DESC;  --  more columns to break ties?

Answer 2

create or replace function searchByTags(tags varchar[])
returns table (id_course integer, name text, quantitiy integer)
as $$
    select *
    from (
        select c.id_course, c.name, count(*) quantity
        from
            course c
            inner join
            tagcourse tc on c.id_course = tc.id_course
            inner join
            unnest(tags) s(tag) on s.tag = tc.tag
        group by c.id_course, c.name
    ) s
    order by quantity desc, name
    ;
$$ language sql;