Or you could wrap your entire SELECT in another query, and do an IFNULL
around the rating field to convert it to 0 (or some other value
important to you) as follows:
SELECT
movie_id,
<... any other fields from movies table you want ...>,
IFNULL(ratings, 0) AS rating
FROM
(
SELECT mo
The ratings field would be NULL. You could also add a count in your
query to tell how many ratings there were. If count is 0, you know
there are no ratings.
SELECT count(ratings.rating_id) AS rate_count, ...
Brent Baisley
On Mon, Dec 22, 2008 at 12:39 PM, Brian Dunning wrote:
> If I did the left
If I did the left join to include movies with no ratings, how would I
tell if it had no ratings? If I used mysql_fetch_array in PHP, would
$result['rating'] == 0, or '', or NULL, or what?
On Dec 22, 2008, at 9:29 AM, Brent Baisley wrote:
The biggest problem is your join condition (and no gr
Pretend I'm Netflix and I want to return a list of found movies,
including the average of related ratings for each movie. Something
like this:
select movies.*, average(ratings.rating) from movies, ratings where
movies.movie_id=ratings.movie_id
I'm sure that's wrong in about 10 different w
