select name,
sum(rating),
group_concat(salary order by salary desc)
from keywords
join salaries on salaries.keywordId = keywords.id
where profession = "backend developer"
group by name
order by sum(rating) DESC
limit 10;
1. git, [ {100k: 10}, {"120k": 16}, {"150k": 5}, {"200k": 2} ]
2. mysql ...
3. ...