簡單的MySQL連接不起作用

Question

我試圖選擇有關其相應投票的評論和摘要統計。並非所有評論都有投票，而那些評論可以有超過1票。這是我的查詢：

select 
    `article-comments`.id, 
    `article-comments`.user as user_id, 
    concat(users.first_name, ' ', users.last_name) as name, 
    users.job_title, 
    users.company_name, 
    avatar, 
    `article-comments`.content, 
    datediff(now(), `article-comments`.date_added) as diff, 
    `article-comments`.date_added as date, 
    count(`comment-votes`.id) as votes_count, 
    sum(`comment-votes`.score) as votes_score 
from 
    `article-comments` left outer join `comment-votes` on `article-comments`.id = `comment-votes`.comment, 
    users 
where 
    `article-comments`.status = 1 
&& `article-comments`.article = $resource_id 
&& `article-comments`.user = users.id 
&& `comment-votes`.status = 1 
order by 
    `article-comments`.id asc 

它沒有連接，但只返回1行與它完美的作品。

任何想法？

Answer 1

嘗試：

select 
       `article-comments`.id, 
       `article-comments`.user as user_id, 
       concat(users.first_name, ' ', users.last_name) as name, 
       users.job_title, 
       users.company_name, 
       avatar, 
       `article-comments`.content, 
       datediff(now(), `article-comments`.date_added) as diff, 
       `article-comments`.date_added as date, 
       count(`comment-votes`.id) as votes_count, 
       sum(`comment-votes`.score) as votes_score 
      from 
       `article-comments` left join `comment-votes` on (`article-comments`.id = `comment-votes`.comment), 
       users 
      where 
       `article-comments`.status = 1 
      && `article-comments`.$table = $resource_id 
      && `article-comments`.user = users.id 
      order by 
       `article-comments`.id asc