Alternative Using Having in mysql - mysql

Question

I am trying to get the records where avg is greater than 81, I noticed I can't use a simple where avg(score) > 80
But using a Having statement is problematic as well as it does not consider where the individual records average is greater than 80, but it considers the group average. Is there an alternative?

Answer 1

In general, if we want to return aggregates (SUM,AVG) and also return detail that makes up the aggregate, we typically use two SELECT
As a rudimentary example, consider a table of "test_score"
test_id student_id score
------- ---------- -----
101 6 90
101 7 71
101 8 88
222 6 93
222 7 78
222 8 81
We can calculate the average score for each test, with a SELECT ... GROUP BY query.
SELECT r.test_id AS test_id
, AVG(r.score) AS avg_score
, MAX(r.score) AS high_score
FROM test_score r
GROUP
BY r.test_id
We expect that to return a resultset like this:
test_id avg_score
------- ---------
101 83
222 84
We can use that query as an inline view i.e. we wrap it in parens and reference it like a table in the FROM clause of another SELECT.
As a demonstration, to return student scores that were better (or equal to) average for each test:
SELECT s.test_id
, s.avg_score
, t.student_id
, t.score
FROM ( -- inline view to get average score for each test_id
SELECT r.test_id AS test_id
, AVG(r.score) AS avg_score
FROM test_score r
GROUP
BY r.test_id
) s
LEFT
JOIN test_score t
ON t.test_id = s.test_id
AND t.score >= s.avg_score
ORDER
BY t.test_id
, s.score DESC
And we'd expect that to return something like:
test_id avg_score student_id score
------- --------- ---------- -----
101 83 6 90
101 83 8 88
222 84 6 93
The first two columns, returned from the inline view, are the result of the aggregate (AVG). The last two columns are detail rows, matched to the rows from the aggregate result.
To summarize the main point here:
To return aggregates along with details, we typically need two SELECT.
One SELECT to get the aggregates (with a GROUP BY if the aggregates are "per" each something or other)
Another SELECT to get the details and a match to the aggregate.

Answer 2

If the average score being computed in your query is already correct, you are just having trouble filtering by it, just wrap it in parens and select from it
select * from (
SELECT Count(entry_id) AS Filled,
q.question AS Questions,
AVG(ag.score) AS TOTAL
FROM entry e
LEFT JOIN entry_answer ea
ON ea.entry_id= e.entry
LEFT JOIN question q
ON q.question_id = ea.question_id
LEFT JOIN question_group qg
ON ea.question_parent_id = qg.question_parent_id
LEFT JOIN answer_group ag
ON ag.question_id = qg.question_parent_id
JOIN sent_list using (sent_list_id)
WHERE
entry_group_id = 2427
AND ag.score >= 0
AND ea.rated_answer_id = ag.rated_answer_id
AND sent_id = 6156
AND e.entry_date BETWEEN '2018-01-01' AND '2019-12-31'
group by ea.question_id
) results where total >= 81