Question

我解释了我的问题。我有这张桌子：

      artist                          artist_pic
-------------------       ---------------------------------------
| id | name | age |       | id_artist | picUrl | picDescription |
-------------------       ---------------------------------------
   |                          /\
   ----------------------------

每位艺术家都有很多照片。

这是我的查询（可行）

SELECT a.*, ai.* 
     FROM artist a INNER JOIN artist_pic ai on ai.id_artist = a.id  
         where ai.id_artist = '" + idArtist + "' and ai.elemento = 'artist'

现在我应该“添加”另一个名为artist_video的表

      artist                          artist_video
-------------------       ---------------------------------------
| id | name | age |       | id_artist | video| videoDescription |
-------------------       ---------------------------------------
   |                          /\
   ----------------------------

我试过这样的事，但我没有得到正确的记录

SELECT a.*, ai.*, av.* 
   FROM artist a 
       INNER JOIN artist_pic ai on ai.id_artist = a.id  
       INNER JOIN artist_video av on av.id_artist = a.id
         where ai.id_artist = 2 and ai.elemento = 'artist' and av.id_artist = 
          2 and av.elemento = 'artist'

例如，我应该得到的是：

michael jackson   Img/5345.jpg     concert      video/5345.avi     concert
michael jackson   Img/3453.jpg     family       video/2344.avi     family

但现在我得到的是（很多次同样的记录）

michael jackson   Img/5345.jpg     concert      video/5345.avi     concert
michael jackson   Img/3453.jpg     family       video/2344.avi     family
michael jackson   Img/5345.jpg     concert      video/5345.avi     concert
michael jackson   Img/3453.jpg     family       video/2344.avi     family
michael jackson   Img/5345.jpg     concert      video/5345.avi     concert
michael jackson   Img/3453.jpg     family       video/2344.avi     family
michael jackson   Img/5345.jpg     concert      video/5345.avi     concert
michael jackson   Img/3453.jpg     family       video/2344.avi     family
michael jackson   Img/5345.jpg     concert      video/5345.avi     concert
michael jackson   Img/3453.jpg     family       video/2344.avi     family

Answer 1

我猜你想要所有的原始唱片，还有那些有视频的唱片。如果是这样，这可能会解决您的问题：

SELECT a.*, ai.*, av.* 
FROM artist a INNER JOIN
     artist_pic ai
     on ai.id_artist = a.id and ai.elemento = 'artist' LEFT JOIN
     artist_video av
     on av.id_artist = a.id and av.elemento = 'artist'
where a.id_artist = 2;

Answer 2

您可以将WHERE条件添加到此查询中，并参阅

SELECT  a.*, ai.*, av.* 
FROM    artist a 
JOIN    artist_pic ai on ai.id_artist = a.id  
JOIN    artist_video av on av.id_artist = a.id
WHERE   a.id = 2

Answer 3

您必须使用UNION命令合并两个查询：

SELECT a.*, 0 as is_video, ai.primarykey
     FROM artist a INNER JOIN artist_pic ai on ai.id_artist = a.id  
         where ai.id_artist = '" + idArtist + "' and ai.elemento = 'artist'
UNION
SELECT a.*, 1 as is_video, av.primarykey
     FROM artist a INNER JOIN artist_video av on av.id_artist = a.id  
         where av.id_artist = '" + idArtist + "' and av.elemento = 'artist'

请注意，它只适用于两个部分具有相同列类型的情况，因此您可能无法使用ai。*和av。*，而是分别指定每个列。我还添加了一个“is_video”列来区分图片和视频（如果需要）。

如果您为artist_picture和artist_video添加名称和类型列，则应该为您提供以下内容：

   michael jackson   Img/5345.jpg     concert
   michael jackson   Img/3453.jpg     family
   michael jackson   video/5345.avi     concert
   michael jackson   video/2344.avi     family

如何用'where'加入2个表

3 个答案: