Question

您好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。我知道还有数以百万计的其他例子，但我无法弄清楚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。我想这是因为应该同时选择concertID和URL。但是不允许SELECT concertID, URL FROM concert WHERE galleryID = g.galleryID ORDER BY RAND() LIMIT 1所以如何解决这个问题？

我在第一个查询中选择的内容是正确的，除了在URL上获取NULL。所以我需要选择的是galleryID和addedDate FROM gallery（每行1个galleryID，而不是相同的8个），concertID和来自Concert的URL（1个随机帖子，具有相同的concertID），名称FROM照片（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;

我还尝试使用此错误#1054 - Unknown column 'p.concertID' in 'where clause'

进行JOIN

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;

Answer 1

在查询中，“FROM”子句中有相关的子查询。事情变得越来越混乱。

看起来你正试图让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”子句来完成此操作：

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

要从8个不同的画廊中获取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

Answer 2

我在SQL中看不出为什么你为URL获取null，他们都有值。

我以其他格式重新发送您的查询，它与您已经执行的操作相似

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

如何将SELECT SQL子查询与RAND组合？

2 个答案: