Question

我有下表：帖子：

const one = 1;
const two = 2;
const result = `One add two is ${one + two}`;
console.log(result); // output: One add two is 3

评论：

id   post
1   'p1'
2   'p2'
3   'p3'

comments_likes：

id   comment post_id
1   'c1'     1
2   'c2'     1
3   'c3'     2

我想查询特定帖子（例如id comment_id like 1 1 1 2 1 1 3 2 -1并获得以下信息：

post_id = 1

我尝试了几次运气不好的查询：例如：

comment_id   comment   likes   
1            'c1'       2
2            'c2'      -1
3            'c3'       0

或

SELECT comment_id, SUM(like) AS likes
FROM comments_likes cl
RIGHT JOIN (
SELECT * 
FROM comments
WHERE post_id=1
) c ON c.id = cl.comment_id;

有人知道如何在单个查询中做到这一点吗？

Answer 1

您可以尝试使用总和

select id, comment, sum( ifnull(like,0) likes  
from  comments 
left  join  comments_likes on a.id = comments_likes.comment_id 
where  post_id =1
group by  id, comment

或针对所有帖子

select post.post,  comments.id, comments.comment, sum( ifnull(commnents_like,0) likes  
from  post
inner join  comments on post.id = comment.post_id 
left  join  comments_likes on a.id = comments_likes.comment_id 

group by   post.post,  comments.id, comments.comment

Answer 2

ifnull(commnents_like,0)

IFNULL（）：如果表达式为NULL，则函数可返回一个替代值：

如何基于另一个SELECT查询的结果求和一列的值？

2 个答案: