有麻烦了从MySQL查询

Question

正确的输出我有包括以下表的数据库：

• 用户
• 类别
• 文章

里面我们的网站，我们有一个部分称为“编辑选择”，其中有5篇编辑在该领域选择的最佳文章。

编辑必须设置“is_recommended = yes”和“recommended_location”，它可以是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“start_date”，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，它应该坐在编辑选择的3盒中。

有时，我们可以具有制品如以下：

• ID：123
• is_recommended：是
• recommended_location = 3
• 起始日期= 2016年6月5日九时00分00秒（假设这是昨天）

其中目前排名第3。

我有另一篇文章：

• ID：456
• is_recommended：是
• recommended_location = 3
• 起始日期= 2016年7月5日九点00分00秒（这是今天和今天已经是上午11点）

但是我的查询仍然显示ID：123;而我希望它显示插槽＃3是最新的（含义456）

有人可以告诉我我做错了什么在我的查询下面，我怎么能保证每个插槽最新的项目是选择？

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

您havent't聚集功能，所以你如果只希望每个recommended_location一个文章，你应该使用

不需要按组（最终使用不同的，如果这是你所需要的）

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 

(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

试试这个：以降序排列查询

使用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。

Answer 3

先进行聚合，然后加入数据，需要

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 ... 

但我f 2或更多文章具有相同的start_date，您将以这种方式获得所有文件...