有麻煩了從MySQL查詢

Question

正確的輸出我有包括以下表的數據庫：

• 用戶
• 類別
• 文章

裏面我們的網站，我們有一個部分稱爲「編輯選擇」，其中有5篇編輯在該領域選擇的最佳文章。

編輯必須設置「is_recommended = yes」和「recommended_location」，它可以是1,2,3,4或5;因此他們將被放置在網站上的1-5個展示位置之一。

文章還有一個「start_date」，意思是作者可以寫一篇文章，將其指定爲is_recommended = yes和recommended_location = 3，然後將其設置爲明天晚上9點。因此這篇文章只會在明天出現，當它出現時，它應該坐在編輯選擇的3盒中。

有時，我們可以具有製品如以下：

• ID：123
• is_recommended：是
• recommended_location = 3
• 起始日期= 2016年6月5日九時00分00秒（假設這是昨天）

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日九點00分00秒（這是今天和今天已經是上午11點）

但是我的查詢仍然顯示ID：123;而我希望它顯示插槽＃3是最新的（含義456）

有人可以告訴我我做錯了什麼在我的查詢下面，我怎麼能保證每個插槽最新的項目是選擇？

這是查詢：

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個recommended_location一個文章，你應該使用

不需要按組（最終使用不同的，如果這是你所需要的）

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

試試這個：以降序排列查詢

使用article.id

所以你的查詢是這樣的：

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 

我希望你能得到解決方案。

Answer 3

先進行聚合，然後加入數據，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，您將以這種方式獲得所有文件...