mysql显示所有行的平均值

Question

Table structure

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

Sample data:

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

Given the structure above, post 1 is rated by user 1, 2 and 3 wherein users 1 and 2 voted for like while user 3 voted for dislike.

With that, how to get the average where rating is a five(5) star rating?

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

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

Rate would be 3.5 out of 5 stars

Expected mysql output would be:

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

Here's what I've done with the sql but I have no idea how to make it into one sql statement only.

This gets the average:

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

In which should be merge into this statement:

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

Merging the two statements gives an error of subquery returns more than 1 row :

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

Adding a where clause on both select statements solves the issue but it is needed to display multiple rows instead.

Answer 1

Nevermind... was able to figure this out. Subquery is used as a joined table instead of a column.

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