Question

我想要的结果只是投票给rateable_id ='3'的用户数量？

示例：正如您在我的表费率中所看到的，rater_id：1和rater_id：8已经投票选出rateable_id = 3.这使得有2个用户。

我的问题是如何在视图中显示？

这是在我的ranking_controller.rb中：

class RankingController < ApplicationController
  def index
    @rankings = Rate.find(:all)
  end
end

这是我的表费率：

- !ruby/object:Rate
  attributes:
    id: 11
    rater_id: 1
    rateable_id: 3
    rateable_type: Bboy
    stars: 5.0
    dimension: foundation
    created_at: 2014-02-25 09:33:23.000000000 Z
    updated_at: 2014-02-25 09:33:23.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 12
    rater_id: 1
    rateable_id: 3
    rateable_type: Bboy
    stars: 5.0
    dimension: originality
    created_at: 2014-02-25 09:33:24.000000000 Z
    updated_at: 2014-02-25 09:33:24.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 13
    rater_id: 1
    rateable_id: 3
    rateable_type: Bboy
    stars: 5.0
    dimension: dynamics
    created_at: 2014-02-25 09:33:25.000000000 Z
    updated_at: 2014-02-25 09:33:25.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 14
    rater_id: 1
    rateable_id: 3
    rateable_type: Bboy
    stars: 5.0
    dimension: execution
    created_at: 2014-02-25 09:33:26.000000000 Z
    updated_at: 2014-02-25 09:33:26.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 15
    rater_id: 1
    rateable_id: 3
    rateable_type: Bboy
    stars: 5.0
    dimension: battle
    created_at: 2014-02-25 09:33:27.000000000 Z
    updated_at: 2014-02-25 09:33:27.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 16
    rater_id: 1
    rateable_id: 5
    rateable_type: Bboy
    stars: 5.0
    dimension: foundation
    created_at: 2014-02-25 09:36:30.000000000 Z
    updated_at: 2014-02-25 09:36:30.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 17
    rater_id: 1
    rateable_id: 5
    rateable_type: Bboy
    stars: 5.0
    dimension: originality
    created_at: 2014-02-25 09:36:31.000000000 Z
    updated_at: 2014-02-25 09:36:31.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 18
    rater_id: 1
    rateable_id: 5
    rateable_type: Bboy
    stars: 5.0
    dimension: dynamics
    created_at: 2014-02-25 09:36:31.000000000 Z
    updated_at: 2014-02-25 09:36:31.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 19
    rater_id: 1
    rateable_id: 5
    rateable_type: Bboy
    stars: 5.0
    dimension: battle
    created_at: 2014-02-25 09:36:32.000000000 Z
    updated_at: 2014-02-25 09:36:32.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 25
    rater_id: 8
    rateable_id: 3
    rateable_type: Bboy
    stars: 1.0
    dimension: foundation
    created_at: 2014-03-04 14:06:46.000000000 Z
    updated_at: 2014-03-04 14:06:46.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 26
    rater_id: 8
    rateable_id: 3
    rateable_type: Bboy
    stars: 1.0
    dimension: originality
    created_at: 2014-03-04 14:06:49.000000000 Z
    updated_at: 2014-03-04 14:06:49.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 27
    rater_id: 8
    rateable_id: 3
    rateable_type: Bboy
    stars: 1.0
    dimension: dynamics
    created_at: 2014-03-04 14:06:51.000000000 Z
    updated_at: 2014-03-04 14:06:51.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 28
    rater_id: 8
    rateable_id: 3
    rateable_type: Bboy
    stars: 1.0
    dimension: execution
    created_at: 2014-03-04 14:06:53.000000000 Z
    updated_at: 2014-03-04 14:06:53.000000000 Z
- !ruby/object:Rate
  attributes:
    id: 29
    rater_id: 8
    rateable_id: 3
    rateable_type: Bboy
    stars: 1.0
    dimension: battle
    created_at: 2014-03-04 14:06:54.000000000 Z
    updated_at: 2014-03-04 14:06:54.000000000 Z

Answer 1

您可以按如下方式使用它：

@raters_count = Rate.select(:rater_id).where(rateable_id: 3, rateable_type: 'Bboy' ).distinct.count

Answer 2

有很多方法可以实现这一目标。

一个是

@ranking_count = Rate.where(rateable_id: 3).pluck(:rater_id).uniq.count

Answer 3

或者这个

@ratings = Rate.group("dimension, rateable_id").select("rateable_id, dimension, count(rater_id) as rate_count")
#=> SELECT rateable_id, dimension,count(rater_id) as rate_count FROM rates GROUP BY dimension, rateable_id

然后

@ratings.each do |rating| 
    rating.dimension
    rating.rateable_id
    rating.rate_count
end

显然可以根据您的需要进行修改，但这样的事情应该足够了。由于您的问题与您的代码不同，因此比原始问题更灵活。您的问题只想处理1 rateable_id，但您的代码会加载所有Rate，因此我根据使用基于维度的Rate方法检索所有rate_count个对象来回答。

更广泛的方法可能是

 @ratings = Rate.group("dimension, rateable_id, rateable_type").select("rateable_id, dimension,rateable_type, count(rater_id) as rate_count, avg(stars) as stars")
  #=> SELECT rateable_id, dimension,rateable_type,count(rater_id) as rate_count, avg(stars) as stars FROM rates GROUP BY dimension, rateable_id,rateable_type

然后

@ratings.each do |rating| 
    rating.dimension
    rating.rateable_id
    rating.rateable_type
    rating.rate_count
    rating.stars
end

根据dimension，rateable_id和rateable_type

分组的费率，您可以访问评分数和平均星数

这应该根据上面的输出（psuedo）返回一些内容

[<Rate dimension:"foundation",rateable_id:3,rateable_type:"Bboy",rate_count:2, stars:3>,
 <Rate dimension:"originality",rateable_id:3,rateable_type:"Bboy",rate_count:2, stars:3>,
 <Rate dimension:"dynamics",rateable_id:3,rateable_type:"Bboy",rate_count:2, stars:3>,
 <Rate dimension:"execution",rateable_id:3,rateable_type:"Bboy",rate_count:2, stars:3>,
 <Rate dimension:"battle",rateable_id:3,rateable_type:"Bboy",rate_count:2, stars:3>,
 <Rate dimension:"foundation",rateable_id:5,rateable_type:"Bboy",rate_count:1, stars:5>,
 <Rate dimension:"originality",rateable_id:5,rateable_type:"Bboy",rate_count:1, stars:5>,
 <Rate dimension:"dynamics",rateable_id:5,rateable_type:"Bboy",rate_count:1, stars:5>,
 <Rate dimension:"battle",rateable_id:5,rateable_type:"Bboy",rate_count:1, stars:5>]

Rails ActiveRecord投票的用户总数？

3 个答案: