Reputation: 6169
CREATE TABLE `reviews` (
`id` int(11) NOT NULL,
`average` decimal(11,2) NOT NULL,
`house_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `reviews` (`id`, `average`, `house_id`) VALUES
(1, '10.00', 1),
(2, '10.00', 1);
ALTER TABLE `reviews`
ADD PRIMARY KEY (`id`);
ALTER TABLE `reviews`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT, AUTO_INCREMENT=3;
CREATE TABLE `dummy_reviews` (
`id` int(11) NOT NULL,
`average` decimal(11,2) NOT NULL,
`house_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `dummy_reviews` (`id`, `average`, `house_id`) VALUES
(0, '2.00', 1);
ALTER TABLE `dummy_reviews`
ADD PRIMARY KEY (`id`);
AND the query
SELECT
AVG(r.average) AS avg1,
AVG(dr.average) AS avg2
FROM
reviews r
LEFT JOIN
dummy_reviews dr ON r.house_id = dr.house_id
the result is
avg1 avg2
10.000000 2.000000
All good by now but (10 + 2) / 2 = 6 ... wrong result
I need (10+10+2) / 3 = 7,33 ... How can I get this result?
Upvotes: 3
Views: 735
Reputation: 1269803
Jorge's answer is the simplest approach (and I duly upvoted it). In response to your comment, you can do the following:
select ( (coalesce(r.suma, 0) + coalesce(d.suma, 0)) /
(coalesce(r.cnt, 0) + coalesce(d.cnt, 0))
) as overall_average
from (select sum(average) as suma, count(*) as cnt
from reviews
) r cross join
(select sum(average) as suma, count(*) as cnt
from dummy_reviews
) d;
Actually, I suggest this not only because of your comment. Under some circumstances, this could be the better performing code.
Upvotes: 3
Reputation: 23361
You have values joined and as such you wont have 3 rows, you will have 2. What you need is a union so you can have all rows from your average tables and do the calculation from it. Like this:
select avg(average) from
(select average from reviews
union all
select average from dummy_reviews
) queries
See it here: http://sqlfiddle.com/#!9/e0b75f/3
Upvotes: 4