1

我有如下表定义:

  • 地点
    (ID,名称)
  • 评论
    (id、userid、placeid)
  • 收藏夹
    (id、userid、placeid)
  • 照片
    (id, url, placeid)
    其中placeid是表 id 的外键Place

在那张桌子上,我想得出这种信息:
- placeid, place name, totalReview, totalFavorite, totalPhoto.

我被卡住了。我目前的进展我可以只从 1 个表中获取信息,就像我可以通过使用这个 mysql 语句知道地方的总评论一样: SELECT p.*, count(r.id) as totalReview from Place p left join Review r on p.id = r.placeid group by p.id。但是,我不知道如何得出totalFavorite 和totalPhoto。

4

2 回答 2

4

您需要分别聚合每个表。这是一种解决方案:

SELECT p.*, 
       totalreview, 
       totalfavorite, 
       totalphoto 
FROM   place p 
       LEFT OUTER JOIN (SELECT placeid, 
                               Count(*) AS totalReview 
                        FROM   review 
                        GROUP  BY placeid) r 
                    ON p.placeid = r.placeid 
       LEFT OUTER JOIN (SELECT placeid, 
                               Count(*) AS totalFavorite 
                        FROM   favorite 
                        GROUP  BY placeid) f 
                    ON p.placeid = f.placeid 
       LEFT OUTER JOIN (SELECT placeid, 
                               Count(*) AS totalPhoto 
                        FROM   photo 
                        GROUP  BY placeid) ph 
                    ON p.placeid = ph.placeid 
于 2012-05-05T03:27:44.893 回答
0

这是一种简单的方法:

SELECT
p.id, p.name,
(SELECT COUNT(*) FROM Review r WHERE r.placeId=p.id) AS totalReview
(SELECT COUNT(*) FROM Favorite f WHERE f.placeId=p.id) AS totalFavorite
(SELECT COUNT(*) FROM Photo ph WHERE ph.placeId=p.id) AS totalPhoto
FROM Place p
于 2012-05-05T03:26:25.660 回答