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嗨，我正在尝试将一些查询合并为一个，但我不确定如何解决这个问题。我知道还有数百万其他示例，但我不知道如何将它们翻译成我的查询。

这是数据库。首先是表名：，然后是主键和外键

gallery:
    galleryID
    name
    addedDate

concert:
    concertID
    galleryID
    name
    URL
    addedDate

photo:
    photoID
    concertID
    name

这是我的查询，它在随机情况下在 URL 上返回 NULL。我猜这是因为应该同时选择 ConcertID 和 URL。但是不允许有SELECT concertID, URL FROM concert WHERE galleryID = g.galleryID ORDER BY RAND() LIMIT 1那么我该如何解决这个问题？

除了在 URL 上获取 NULL 之外，我在第一个查询中选择的内容是正确的。所以我需要选择的是galleryID和addDate FROM gallery（每行1个galleryID不是8个相同），concertID和来自音乐会的URL（1个具有相同concertID的随机帖子），来自照片的名称（1个具有相同concertID的随机帖子）。给我这些结果：

11  2012-07-31 15:44:35 90  Picture\Path11  SomePicture28.jpg
36  2012-07-31 14:31:36 208 Picture\Path36  SomePicture11.jpg
09  2012-07-30 15:28:02 33  Picture\Path09  SomePicture69.jpg

SELECT galleryID, addedDate, 
    (SELECT concertID 
        FROM concert 
            WHERE galleryID = g.galleryID 
    ORDER BY RAND() LIMIT 1) AS curID, 
    (SELECT URL 
        FROM concert 
            WHERE concertID = curID) AS URL, 
    (SELECT p.name 
            FROM photo p, concert c 
                WHERE p.concertID = curID AND c.galleryID = g.galleryID 
     ORDER BY RAND() LIMIT 1) AS photoName 
FROM gallery g ORDER BY addedDate DESC LIMIT 8;

我也尝试通过 JOIN 来执行此错误#1054 - Unknown column 'p.concertID' in 'where clause'

SELECT galleryID, addedDate, c.concertID, c.URL, p.name 
    FROM (SELECT concertID, URL, 
            (SELECT name 
                FROM photo 
                    WHERE p.concertID = curID.concertID 
             ORDER BY RAND() LIMIT 1) AS photoName 
             FROM concert 
                WHERE c.galleryID = curID.galleryID 
          ORDER BY RAND() LIMIT 1) curID 
LEFT JOIN concert c ON curID.galleryID = c.galleryID 
LEFT JOIN photo p ON p.name = curID.photoName 
ORDER BY addedDate DESC LIMIT 8;
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2 回答 2

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我在您的 SQL 中看不到任何原因为什么您的 URL 为空，它们是否都有值。

我已经用另一种格式重新完成了您的查询，这与您已经做的类似

SELECT R.*, c.URL, (SELECT p.Name FROM Photo p WHERE p.ConcertID = R.CurID ORDER BY RAND() LIMIT 1) AS PhotoName
FROM 
(SELECT GalleryID, AddedDate,(SELECT ConcertID FROM Concert c WHERE c.GalleryID = g.GalleryID ORDER BY RAND() LIMIT 1) AS CurID
FROM Gallery g
ORDER BY AddedDate DESC LIMIT 8) AS R
JOIN Concert c ON R.CurID = c.ConcertID
于 2012-08-03T23:24:29.603 回答
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在您的查询中，您在“FROM”子句中有一个相关的子查询。从那里开始，事情变得越来越混乱。

看起来您正在尝试获取一张与音乐会相关的照片（在画廊中）。您可以通过将相关子查询移动到“SELECT”子句来做到这一点：

SELECT galleryID, addedDate, c.concertID, c.URL,
       (select name
        from photo p
        where p.concertID = c.concertId
        order by rand()
        limit 1
       ) photoname
FROM concert c left join
      gallery g
      ON g.galleryID = c.galleryID
ORDER BY addedDate DESC
LIMIT 8;

要从 8 个不同的画廊获取 8 张照片，请尝试以下操作：

SELECT galleryID, addedDate, c.concertID, c.URL,
       (select name
        from photo p
        where p.concertID = c.concertId
        order by rand()
        limit 1
       ) photoname
FROM (select g.*
      from gallery g
      order by rand()
      limit 8
     ) g join
     concert c
     ON g.galleryID = c.galleryID
where c.concert_id =
          (select csub.concert_id
           from concert csub
           where csub.galleryID = g.galleryID
           order by rand()
           limit 1
          ) 
ORDER BY addedDate DESC
于 2012-08-02T18:22:27.703 回答