Question

我有一个包含以下表格的数据库：

用户
分类
文章

在我们的网站内，我们有一个名为＆＃34;编辑选择＆＃34;这个领域的编辑选择了5篇最好的文章。

编辑必须设置＆＃34; is_recommended = yes＆＃34;以及＆＃34; recommended_location＆＃34;可以是1,2,3,4或5;因此，他们将被放置在网站上的1-5个展示位置之一。

文章还有一个＆＃34; start_date＆＃34;意思是作者可以写一篇文章，将其指定为is_recommended = yes和recommended_location = 3，然后将其设置为明天晚上9点。因此，这篇文章只会在明天出现，当它出现时，它应该位于编辑选择的3个方框中。

有时我们可能会有如下文章：

ID：123
is_recommended：是
recommended_location = 3
start_date = 06-05-2016 09:00:00（让我们说这是昨天）

目前正处于第3位。

我还有另一篇文章：

ID：456
is_recommended：是
recommended_location = 3
start_date = 07-05-2016 09:00:00（这是今天，今天已经是上午11点）

但是我的查询仍然显示ID：123;虽然我希望它显示插槽＃3中最新的（意思是456）

有人可以在下面的查询中告诉我我做错了什么，我怎样才能确保每个插槽都选择了最新的项目？

这是查询：

select * 
from (
    select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
    from article 
    left join user on article.created_by = user.id 
    left join category on category.id = article.category_id 
    where article.status='active' 
    AND is_recommended='yes' 
    AND article.start_date<='".date('Y-m-d H:i:s')."' 
    AND recommended_location in (1,2,3,4,5) 
    order by start_date desc
 ) as x 
 group by recommended_location 
 limit 5

Answer 1

你没有聚合功能所以你不需要分组（如果这是你需要的话，最终会使用不同的）

select * 
from (    select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
    from article 
    left join user on article.created_by = user.id 
    left join category on category.id = article.category_id 
    where article.status='active' 
    AND is_recommended='yes' 
    AND article.start_date<='".date('Y-m-d H:i:s')."' 
    AND recommended_location in (1,2,3,4,5) 
    order by start_date desc
 ) as x 
 limit 5

如果您只想为每个recommended_location添加一篇文章，则应使用

(select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
from article 
left join user on article.created_by = user.id 
left join category on category.id = article.category_id 
where article.status='active' 
AND is_recommended='yes' 
AND article.start_date<='".date('Y-m-d H:i:s')."' 
AND recommended_location = '1' 
order by start_date desc limit 1)
union
(select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
from article 
left join user on article.created_by = user.id 
left join category on category.id = article.category_id 
where article.status='active' 
AND is_recommended='yes' 
AND article.start_date<='".date('Y-m-d H:i:s')."' 
AND recommended_location = '2' 
order by start_date desc limit 1)
union 
(select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
from article 
left join user on article.created_by = user.id 
left join category on category.id = article.category_id 
where article.status='active' 
AND is_recommended='yes' 
AND article.start_date<='".date('Y-m-d H:i:s')."' 
AND recommended_location = '3' 
order by start_date desc limit 1)   
union 
(select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
from article 
left join user on article.created_by = user.id 
left join category on category.id = article.category_id 
where article.status='active' 
AND is_recommended='yes' 
AND article.start_date<='".date('Y-m-d H:i:s')."' 
AND recommended_location = '4' 
order by start_date desc limit 1)  
union 
(select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
from article 
left join user on article.created_by = user.id 
left join category on category.id = article.category_id 
where article.status='active' 
AND is_recommended='yes' 
AND article.start_date<='".date('Y-m-d H:i:s')."' 
AND recommended_location = '5' 
order by start_date desc limit 1)

Answer 2

首先进行聚合，然后加入所需的数据

select x.recommended_location, x.start_date, ... 
from
 ( select article.recommended_location, max(article.start_date) as  start_date 
    from article 
     where article.status='active' 
     AND is_recommended='yes' 
     AND article.start_date<='".date('Y-m-d H:i:s')."' 
    AND recommended_location in (1,2,3,4,5) 
    group by article.recommended_location
 ) as x 
inner join article on x.recommended_location = artice.recommended_location    
and x.start_date = article.start_date
inner join ...

但是，如果2篇或更多文章具有相同的start_date，您将以这种方式获得所有这些...

Answer 3

试试这个：

在查询中使用降序的article.id

所以你的查询是这样的：

select * 
from (
    select article.*, user.username, category.title as ctitle, user.firstname, user.lastname, category.slug as cslug, category.category_id as pid 
    from article 
    left join user on article.created_by = user.id 
    left join category on category.id = article.category_id 
    where article.status='active' 
    AND is_recommended='yes' 
    AND article.start_date<='".date('Y-m-d H:i:s')."' 
    AND recommended_location in (1,2,3,4,5) 
    order by article.ID desc, recommended_location desc, start_date desc
 ) as x 
 group by recommended_location 
 limit 5

我希望你能得到解决方案。

无法从MySQL查询中获得正确的输出

3 个答案: