Question

我有桌子'海报'，表'评论'和表'预告片'。每个表都与一个movieID列相关联，但对于某些movieID，每个表都可以为空：

 ++ posters_table +++      ++ reviews_table ++      ++ trailers_table ++

--itemID--+--filename-    --itemID--+--review--    --itemID--+--trailer
----------------------    ---------------------    ---------------------
----001---+--0012343--    ----004---+--blalba--    ----002---+--002345--
----001---+--0013331--    ----004---+--xlalxa--    ----005---+--005434--
----002---+--0020052--    ----005---+--zlalza--    ----001---+--005335--

我想COUNT（）指定movieID的海报，评论和预告片的数量，如果没有，则获得0。

所以，如果我想算上movieID = 001，我得到：['posters'] = 2 / ['reviews'] = 0和['trailers'] = 1（例如）

有人可以发布SQL查询来执行此操作吗？

Answer 1

select
    (select count(*) from posters_table  where itemId = ?) as posters,
    (select count(*) from reviews_table  where itemId = ?) as reviews,
    (select count(*) from trailers_table where itemId = ?) as trailers;

Answer 2

我认为如果您将影片表格连接到给出计数的结果集，则可以选择计数是否为＆gt; 0并给出适当的值：

SELECT movieID, IF(posters.posters_count > 0, posters.posters_count,0) AS posters_total,  IF(reviews.reviews_count > 0, reviews.reviews_count,0) AS reviews_total,  IF(trailers.trailers_count > 0, trailers.trailers_count,0) AS trailers_total
FROM movies m
LEFT JOIN (SELECT itemID,COUNT(*) AS posters_count FROM posters_table WHERE itemID = '001' GROUP BY itemID) posters ON posters.itemID = movies.movieID
LEFT JOIN (SELECT itemID,COUNT(*) AS reviews_count FROM reviews_table WHERE itemID = '001' GROUP BY itemID) reviews ON reviews.itemID = movies.movieID
LEFT JOIN (SELECT itemID,COUNT(*) AS trailers_count FROM trailers_table WHERE itemID = '001' GROUP BY itemID) trailers ON trailers.itemID = movies.movieID
WHERE m.movieID = '001'

编辑：我更喜欢 ar 的解决方案。更简单！

Answer 3

select 
    (select count(P.movieid) from posters P where P.movieid=1),
    (select count(R.movieid) from reviews R where R.movieid=1),
    (select count(T.movieid) from trailers T where T.movieid=1)

这些表并非“真正”加入，因此您需要三个选择。

MySQL：OUTER JOIN和COUNT（）

3 个答案: