Question

表格结构

rating.post_id
rating.user_id
rating.rate_like
rating.rate_dislike

示例数据：

rating.post_id = 1
rating.user_id = 1
rating.rate_like = 1
rating.rate_dislike = 0

rating.post_id = 1
rating.user_id = 2
rating.rate_like = 1
rating.rate_dislike = 0

rating.post_id = 1
rating.user_id = 3
rating.rate_like = 0
rating.rate_dislike = 1

鉴于上述结构，帖子1由用户1,2和3评级，其中用户1和2投票赞成用户3投票不喜欢。

有了这个，如何得到评级为五（5）星评级的平均值？

式： (((total rate_like / total votes) * 100) * 0.5) / 10 = Rate (round to nearest .5)

实施例： (((2/3) * 100) * 0.5) / 10 = 3.33 (round to 3.5)

评分为3.5分，满分5分

预期的mysql输出将是：

post_id = 1
rate_like = 2
rate_dislike = 1
total_rating = 3.5

这是我对sql所做的事情，但我不知道如何只将它变成一个sql语句。

这得到了平均值：

select round(round((((sum(rate_like) / (sum(rate_like) + sum(rate_dislike)) * 100) * 0.5) / 10) * 2) / 2, 1) from rating group by post_id

其中应合并到此声明中：

SELECT post_id, sum(rate_like), sum(rate_dislike), (sum(rate_like) + sum(rate_dislike)) from rating group by post_id

合并这两个语句会出现subquery returns more than 1 row错误：

SELECT post_id, sum(rate_like), sum(rate_dislike), (sum(rate_like) + sum(rate_dislike)), (select round(round((((sum(rate_like) / (sum(rate_like) + sum(rate_dislike)) * 100) * 0.5) / 10) * 2) / 2, 1) from rating group by post_id) from rating group by post_id

在where个语句中添加select子句可以解决问题，但需要显示多行。

Answer 1

没关系......能够解决这个问题。子查询用作连接表而不是列。

SELECT rating.post_id, sum(rate_like), sum(rate_dislike), (sum(rate_like) + sum(rate_dislike)), column_rating from rating inner join ( select post_id, round(round((((sum(rate_like) / (sum(rate_like) + sum(rate_dislike)) * 100) * 0.5) / 10) * 2) / 2, 1) as column_rating from rating group by post_id ) as rating_table on rating.post_id = rating_table.post_id group by rating.post_id

mysql显示所有行的平均值

1 个答案: