Question

我有声明：

$statement_q = "
SELECT img 
     , title
     , time
     , videos.id 
     , date 
  FROM videos 
  JOIN videocategories 
    ON videos.id = videocategories.video_id 
  JOIN categories 
    ON categories.id = videocategories.category_id 
 WHERE categories.name != 'Western'
 GROUP 
    BY videos.id 
 ORDER 
    BY videos.id 
 LIMIT {$startpoint} , {$per_page}";

层次：

如果电影有多个类别，它会显示它！

电影1：Werstern，Adventure（它会显示）

电影2：西方（一类，所以不会显示）

如何排除西方电影？

Answer 1

这样的事情应该有效。我现在无法创建表格来验证它。

$statement_q = "SELECT img,title,time,videos.id as id,date FROM videos 
INNER JOIN videocategories ON videos.id = videocategories.video_id 
INNER JOIN categories on categories.id = videocategories.category_id 
WHERE categories.name != 'Western'
AND not exists (select v.id from videocategories v where v.video_id = videos.id
                and v.category_id <> videocategories.category_id)
GROUP by videos.id 
ORDER BY videos.id 
LIMIT {$startpoint} , {$per_page}";

Answer 2

DROP TABLE IF EXISTS video_category;

CREATE TABLE video_category
(video_id INT NOT NULL
,category_id INT NOT NULL
,PRIMARY KEY(video_id,category_id)
);

INSERT INTO video_category VALUES
(101,1),(101,2),(102,1),(102,3);

SELECT * FROM video_category;
+----------+-------------+
| video_id | category_id |
+----------+-------------+
|      101 |           1 |
|      101 |           2 |
|      102 |           1 |
|      102 |           3 |
+----------+-------------+

SELECT * 
  FROM video_category x 
  LEFT 
  JOIN video_category y 
    ON y.video_id = x.video_id 
   AND y.category_id = 2;
+----------+-------------+----------+-------------+
| video_id | category_id | video_id | category_id |
+----------+-------------+----------+-------------+
|      101 |           1 |      101 |           2 |
|      101 |           2 |      101 |           2 |
|      102 |           1 |     NULL |        NULL |
|      102 |           3 |     NULL |        NULL |
+----------+-------------+----------+-------------+

SELECT DISTINCT x.video_id 
           FROM video_category x 
           LEFT 
           JOIN video_category y 
             ON y.video_id = x.video_id 
            AND y.category_id = 2 
          WHERE y.category_id IS NULL;
+----------+
| video_id |
+----------+
|      102 |
+----------+

如何按类别排除记录

2 个答案: