Question

我试图创建一个MYSQL查询来选择最新的10个广告记录，其中图像的数量大于0.问题是我在'where子句'中得到错误“未知列'num_photos'”因为where语句在计数之前执行。我将如何重写这个mysql查询以解决这个问题？

    
SELECT adverts.*,
( SELECT advert_images.imagethumbpath FROM advert_images WHERE advert_images.advert_id = adverts.advert_id  ORDER BY image_id ASC LIMIT 1 ) as imagethumbpath,
( SELECT count(advert_images.advert_id) from advert_images WHERE advert_images.advert_id = adverts.advert_id ) AS num_photos
FROM adverts
WHERE num_photos > 0
ORDER BY date_updated DESC 
LIMIT 10

Answer 1

您可以使用HAVING代替WHERE

来执行此操作

SELECT adverts.*,
( SELECT advert_images.imagethumbpath FROM advert_images WHERE advert_images.advert_id = adverts.advert_id  ORDER BY image_id ASC LIMIT 1 ) as imagethumbpath,
( SELECT count(advert_images.advert_id) from advert_images WHERE advert_images.advert_id = adverts.advert_id ) AS num_photos
FROM adverts
HAVING num_photos > 0
ORDER BY date_updated DESC 
LIMIT 10

Answer 2

你必须使用having子句而不是where

having num_photos > 0

Answer 3

如果您的大部分广告都有图片，那么这应该非常快。

select a.advert_id
      ,a.date_updated
      ,b.imagethumbpath
  from adverts        a
  join advert_images  b on(a.advert_id = b.advert_id)
 where b.image_id = 
        (select min(c.image_id)
           from advert_image c
          where c.advert_id = b.advert_id)
 order 
    by date_updated desc
 limit 10;

您应该对adverts(date_updated)和advert_images(advert_id, image_id)进行索引。

让我知道这是如何解决的。

使用MYSQL的where子句中的图像计数

3 个答案: