Answer to question
COALESCE
would be utterly pointless where you placed it. The aggregate function count()
never returns NULL
. And logic dictates that num_60_79
can never be lower than 1 in the subquery. Related:
But that's irrelevant. A subquery returning no row cannot return any values at all.
To be precise, in your case the subquery may return any number of rows but, when joining to second_table
, if there is no row matching on (assignment_id, num_60_79)
then all columns from the subquery b
are filled with NULL
values instead. See:
NATURAL FULL JOIN
is a very exotic way to join tables. Especially for somebody still learning how to use COALESCE
. The manual:
NATURAL
is shorthand for a USING
list that mentions all columns in
the two tables that have the same names.
And:
FULL OUTER JOIN
returns all the joined rows, plus one row for each
unmatched left-hand row (extended with nulls on the right), plus one
row for each unmatched right-hand row (extended with nulls on the left).
I think it's safe to assume there is no column named num_60_79
in second_table
, and you surely wouldn't want to involve it in the join conditions if there was one. You only want to join on the column assignment_id
, and that's what you should put in the query instead of the NATURAL
key word.
Also, while FULL JOIN
is theoretically possible here, we would typically see a LEFT JOIN
.
You also don't need two layers of subqueries and some other noise in the query.
Proper query
All things considered, while you are ...
trying to count the number of columns that is retrieved from the subquery
... my educated guess is you want this query instead:
SELECT s.*, COALESCE(b.num_60_79, 0) AS num_60_79
FROM second_table s
LEFT JOIN (
SELECT assignment_id, count(*) AS num_60_79
FROM avg_required_table
WHERE mark_as_percent >= 60
AND mark_as_percent < 80
GROUP BY assignment_id
) b USING (assignment_id);