Sql conditional count with join - mysql

Question

I cannot find the answer to my problem here on stackoverflow. I have a query that spans 3 tables:
newsitem
+------+----------+----------+----------+--------+----------+
| Guid | Supplier | LastEdit | ShowDate | Title | Contents |
+------+----------+----------+----------+--------+----------+
newsrating
+----+----------+--------+--------+
| Id | NewsGuid | UserId | Rating |
+----+----------+--------+--------+
usernews
+----+----------+--------+----------+
| Id | NewsGuid | UserId | ReadDate |
+----+----------+--------+----------+
Newsitem obviously contains newsitems, newsrating contains ratings that users give to newsitems, and usernews contains the date when a user has read a newsitem.
In my query I want to get every newsitem, including the number of ratings for that newsitem and the average rating, and how many times that newsitem has been read by the current user.
What I have so far is:
select newsitem.guid, supplier, count(newsrating.id) as numberofratings,
avg(newsrating.rating) as rating,
count(case usernews.UserId when 3 then 1 else null end) as numberofreads from newsitem
left join newsrating on newsitem.guid = newsrating.newsguid
left join usernews on newsitem.guid = usernews.newsguid
group by newsitem.guid
I have created an sql fiddle here: http://sqlfiddle.com/#!9/c8add/8
Both count() calls don't return the numbers I want. numberofratings should return the total number of ratings for that newsitem (by all users). numberofreads should return the number of reads for the current user for that newsitem.
So, newsitem with guid d104c330-c319-40e8-8be3-a7c4f549d35c should have 2 ratings and 3 reads for the current user with userid = 3.
I have tried conditional counts and sums, but no success yet. How can this be accomplished?

Answer 1

The main problem that I see is that you're joining in both tables together, which means that you're going to effectively be multiplying out by both numbers, which is why your counts aren't going to be correct. For example, if the Newsitem has been read 3 times by the user and rated by 8 users then you're going to end up getting 24 rows, so it will look like it has been rated 24 times. You can add a DISTINCT to your COUNT of the ratings IDs and that should correct that issue. Average should be unaffected because the average of 1 and 2 is the same as the average of 1, 1, 2, & 2 (for example).
You can then handle the reads by adding the userid to the JOIN condition (since it's an OUTER JOIN it shouldn't cause any loss of results) instead of in a CASE statement for your COUNT, then you can do a COUNT on distinct id values from Usernews. The resulting query would be:
SELECT
I.guid,
I.supplier,
COUNT(DISTINCT R.id) AS number_of_ratings,
AVG(R.rating) AS avg_rating,
COUNT(DISTINCT UN.id) AS number_of_reads
FROM
NewsItem I
LEFT OUTER JOIN NewsRating R ON R.newsguid = I.guid
LEFT OUTER JOIN UserNews UN ON
UN.newsguid = I.guid AND
UN.userid = #userid
GROUP BY
I.guid,
I.supplier
While that should work, you might get better results from a subquery, as the above needs to explode out the results and then aggregate them, perhaps unnecessarily. Also, some people might find the below to be a little clearer.
SELECT
I.guid,
I.supplier,
R.number_of_ratings,
R.avg_rating,
COUNT(*) AS number_of_reads
FROM
NewsItem I
LEFT OUTER JOIN
(
SELECT
newsguid,
COUNT(*) AS number_of_ratings,
AVG(rating) AS avg_rating
FROM
NewsRating
GROUP BY
newsguid
) R ON R.newsguid = I.guid
LEFT OUTER JOIN UserNews UN ON UN.newsguid = I.guid AND UN.userid = #userid
GROUP BY
I.guid,
I.supplier,
R.number_of_ratings,
R.avg_rating

Answer 2

I'm with Tom you should use a subquery to calculate the user count.
SQL Fiddle Demo
SELECT NI.guid,
NI.supplier,
COUNT(NR.ID) as numberofratings,
AVG(NR.rating) as rating,
user_read as numberofreads
FROM newsitem NI
LEFT JOIN newsrating NR
ON NI.guid = NR.newsguid
LEFT JOIN (SELECT NewsGuid, COUNT(*) user_read
FROM usernews
WHERE UserId = 3 -- use a variable #user_id here
GROUP BY NewsGuid) UR
ON NI.guid = UR.NewsGuid
GROUP BY NI.guid,
NI.supplier,
numberofreads;