SQL：按组汇总查询结果

Question

I just started learning SQL a few days ago and I am working with some basic examples. In this example I want to find trainings that my friends and teammates like (based on how many of my friends liked the training, how many friends completed the training, and the average rating from my friends on the training). So far I used sub-queries to get me a weighted value for each type of weighted value per training. I want to sum the weighted values for each training giving each training a total and then order the training list by the total and select the top few from that.

Here is what I did.

SELECT training_id, (5* COUNT()) AS [value]FROM progress
WHERE user_id IN (SELECT friend_id AS user_ids FROM friendships WHERE user_id=6 UNION SELECT user_id FROM membership WHERE team_id IN (SELECT team_id FROM membership WHERE user_id = 6))
AND completed = 1
GROUP BY training_id

UNION ALL

SELECT training_id, (10 * AVG(rating))
FROM reviews
WHERE user_id IN (SELECT friend_id AS user_ids FROM friendships WHERE user_id=6 UNION SELECT user_id FROM membership WHERE team_id IN (SELECT team_id FROM membership WHERE user_id = 6))
GROUP BY training_id

UNION ALL

SELECT training_id, COUNT()
FROM likes
WHERE user_id IN (SELECT friend_id AS user_ids FROM friendships WHERE user_id=6 UNION SELECT user_id FROM membership WHERE team_id IN (SELECT team_id FROM membership WHERE user_id = 6))
GROUP BY training_id

The results look like:

training_id     value
______________________
1                10
2                5
1                34.5
2                45
1                6
2                3

Please let me know if you know of a way to do this or if there is an alternative to this method that I should be looking at.

(the tables 'likes' 'reviews' and 'progress' all contain the field "training_id". What do you think about Joining the three tables to begin with?)

Answer 1

This is quite impressive. Your query is well structured and easy to read. If you want sums per training_id, simply build an outer query upon your own:

select training_id, sum(value)
from (<your query here>) x
group by training_id
order by sum(value) desc;

As to joining the tables: No, generally you don't want to do this. You don't want to join each progress record with each review record and each likes record per training_id. This would multiply your values (eg 2 progress entries, 3 reviews, and 4 likes for one training would give you 2 x 3 x 4 = 24 intermediate records). In your specific case this would do no harm, as AVG would result in the same value and you'd replace COUNT(*) with COUNT(DISTINCT progress_id) and COUNT(DISTINCT likes_id) , but once you'd use SUM you're in trouble.

What you could do is join the aggregates:

with mates as
(
  select friend_id as user_id from friendships where user_id = 6
  union 
  select user_id from membership where team_id in 
               (select team_id from membership where user_id = 6)
)
select 
  training_id, 
  coalesce(p.value, 0) + 
  coalesce(r.value, 0) + 
  coalesce(l.value, 0) as total
from
(
  select training_id, 5 * count(*) as value
  from progress
  where user_id in (select user_id from mates)
  and completed = 1
  group by training_id
) p
full outer join
(
  select training_id, 10 * avg(rating) as value
  from reviews
  where user_id in (select user_id from mates)
  group by training_id
) r using (training_id)
full outer join
(
  select training_id, count(*) as value
  from likes
  where user_id in (select user_id from mates)
  group by training_id
) l using (training_id)
order by 2 desc;

Answer 2

OK, thank you for your input. I was right on my second hunch. I needed to use Joins. I only read about Joins in the documentation and wasn't familiar with them, but once I tried it, it was pretty straight forward.

Here is the solution.

SELECT training.*, ((10 * AVG(reviews.rating)) + COUNT(DISTINCT likes.user_id) + ( 5* COUNT(DISTINCT progress.trainee_id))) AS total FROM training JOIN likes ON training.training_id = likes.training_id JOIN reviews ON training.training_id = reviews.training_id JOIN progress ON training.training_id = progress.training_id WHERE user_id IN (SELECT friend_id AS user_ids FROM friendships WHERE user_id=6 UNION SELECT user_id FROM membership WHERE team_id IN (SELECT team_id FROM membership WHERE user_id = 6)) AND progress.completed = 1 AND progress.trainee_id IN (SELECT friend_id AS user_ids FROM friendships WHERE user_id=6 UNION SELECT user_id FROM membership WHERE team_id IN (SELECT team_id FROM membership WHERE user_id = 6)) ORDER BY total DESC GROUP BY training_name LIMIT 6

This gave me
training_id, training_name, total --these are the field names of the result
1, Beginner LiveCode, 53
2, SQL 101, 48