mysql - sql语句的接受率

Question

我有一个具有以下结构的“建议”表：

id(suggest_id) | author_id | accepted(true/false)

我想按最大接受率订购，例如：

Jack had 10 suggests and he accepted 5 (50% acceptance rate)
John had 20 suggests and he accepted 5 (25% acceptance rate)
Steve had 10 suggests and he accepted 8 (80% acceptance rate)

这将返回：史蒂夫、杰克和约翰。

我认为可能必须使用两个 SQL 查询，一个用于建议数量，第二个用于accepted=true.

也许可以通过一个查询来完成？

我正在使用rails，所以它也可以通过rails来完成。

Answer 1

取决于你如何表示接受（真/假），比如......

select author_id, 
    sum(case when accepted ='true' then 1 else 0 end), 
    100.0*sum(case when accepted ='true' then 1 else 0  end)/count(*)
from yourtable
group by author_id
order by 100.0*sum(case when accepted ='true' then 1 else 0 end)/count(*) desc

Answer 2

你也可以尝试使用纯红宝石：

@authors_by_acceptance_rate = Array.new
@authors = Author.all
@authors.each do |author|
  @suggests = Suggest.find_by_author_id(author.id)
      accepted = 0
  @suggests.each do |suggest|
    if suggest.accepted
      accepted = accepted + 1
    end
  end
  acceptance_rate = accepted/@suggests.count
  @authors_by_acceptance_rate.push('name' => @author.name, 'accepted' =>     acceptance_rate)   
end
return @authors_by_acceptance_rate

我还没有测试过这个，所以不能保证它的工作...