Question

我正在开发像Stackoverflow或Disqus这样的评论系统，人们可以对评论进行评论和投票。我有三个表：USERS，COMMENTS和VOTES。

我无法弄清楚如何编写查询以计算投票数并返回给定用户是否投票。每个用户每条评论只能投票一次，不能对自己的评论进行投票。此外，评论和投票不只是一个主题。每个页面都有自己的主题，因此在进行GET到SELECT评论和投票时，WHERE topic_id='X'也需要反映在查询中。

这是我的表格数据，我的查询到目前为止的样子，以及我希望它返回的内容：

USERS table
user_id user_name   
 1         tim
 2         sue
 3         bill 
 4         karen
 5         ed

COMMENTS table
comment_id topic_id    comment   commenter_id
 1            1       good job!         1
 2            2       nice work         2
 3            1       bad job :)        3

VOTES table
 vote_id    vote  comment_id  voter_id
  1          -1       1          5
  2           1       1          4
  3           1       3          1
  4          -1       2          5
  5           1       2          4

SELECT users.*, comments.*, count(vote as totals), sum(vote=1 as yes), sum(vote=-1 as no),
my_votes.vote as did_i_vote, users.* as IM_THE_USER
from comments
join votes 
on comments.comment_id=votes.comment_id
join users
on comments.commenter_id=users.user_id
join votes as MY_votes
on MY_votes.voter_id=IM_THE_USER.user_id
where topic_id=1 and IM_THE_USER.user_id=1
group by comment_id

user_id user_name comment_id topic_id  comment   commenter_id  totals yes no did_i_vote
   1      tim          1        1      good job!     1           2     1   1    NULL  
   3      bill         3        1      bad job:)     3           1     1   0      1

Answer 1

SQL Fiddle播放

select u.user_id, u.user_name,
       c.comment_id, c.topic_id,
       sum(v.vote) as totals, sum(v.vote > 0) as yes, sum(v.vote < 0) as no,
       my_votes.vote as did_i_vote
from comments c
join users u on u.user_id = c.commenter_id
left join votes v on v.comment_id = c.comment_id
left join votes my_votes on my_votes.comment_id = c.comment_id
                            and my_votes.voter_id = 1
where c.topic_id = 1
group by c.comment_id, u.user_name, c.comment_id, c.topic_id, did_i_vote;

更新：固定分组

通过OP（tim peterson）更新

：固定计数为总计并添加注释本身以供查询返回 final working SQLFiddle

投票SQL查询的类似Stackoverflow的评论

1 个答案: