将包括如何获取最多点击记录和至少一个子级别资源

Question

我有这样的具有多态关系的模型

class Level1
 has_and_belongs_to_many :level2s
 has_many :resources ,:as => :mediable
end

class Level2
    has_and_belongs_to_many :level1s
    has_many :level3s
    has_many :resources ,:as => :mediable
end

class Level3
    belongs_to :level2
    has_many :resources ,:as => :mediable
end

class Resource
    belongs_to :mediable , polymorphic: true
    has_many :resources ,:as => :mediable
    has_many :clicks ,:as => :mediable
end

class click
    belongs_to :clickable , polymorphic: true
end

当用户在 level1/level2/level3（图像或媒体）中添加资源时，我将这些媒体显示在用户可以点击的地方，每次点击我都会在点击表中保存一个条目

现在我需要当用户在 level1 的显示页面上时，我需要根据点击次数显示 level1s 和 level2s 组合的前 50 个资源，并且将从数据库中获取至少一个资源

我会这样尝试：

Resource.select("resources.*, count(clicks.id) as click_counts")
            .joins( "INNER JOIN clicks ON clicks.clickable_id = resources.id AND clicks.clickable_type='Resource'" )
            .where("(resources.mediable_id IN(1) AND resources.mediable_type='Level1') OR (resources.mediable_id IN(1, 2, 3, 4, 5) AND resources.mediable_type='Level2')")
            .group("resources.id")
            .order("click_counts").limit(50)

它将返回与级别 1 及其相关级别 2 相关的前 50 个资源，但不保证我至少有一个与级别 2 相关的资源。

你能帮我怎么做吗

有可能从未点击过资源，但我必须获得该资源，并且每个级别至少需要一个资源 所以我认为内连接应该改为左外连接

Answer 1

如果你做一些额外的计算对你来说不是问题，你可以通过以下方式轻松实现：

resources = Resource.select("resources.*, count(clicks.id) as click_counts")
            .joins( "INNER JOIN clicks ON clicks.clickable_id = resources.id AND clicks.clickable_type='Resource'" )
            .where("(resources.mediable_id IN(1) AND resources.mediable_type='Level1') OR (resources.mediable_id IN(1, 2, 3, 4, 5) AND resources.mediable_type='Level2')")
            .group("resources.id")
            .order("click_counts").limit(50)

unless resources.pluck(:mediable_type).include? 'Level2'

  resources = resources.limit(49) + Resource.select("resources.*, count(clicks.id) as click_counts")
            .joins( "INNER JOIN clicks ON clicks.clickable_id = resources.id AND clicks.clickable_type='Resource'" )
            .where("resources.mediable_id IN(1, 2, 3, 4, 5) AND resources.mediable_type='Level2'")
            .group("resources.id")
            .order("click_counts").limit(1)
end

否则，您可以尝试这样的方法（未测试）：

sub_select = "(
   SELECT resources.*, count(clicks.id) as click_counts, 1 as SortKey
   FROM resources 
   INNER JOIN clicks ON clicks.clickable_id = resources.id AND clicks.clickable_type='Resource'
   WHERE resources.mediable_id IN(1, 2, 3, 4, 5) AND resources.mediable_type='Level2'
   GROUP BY resources.id
   ORDER BY mediable_type, click_counts 
   LIMIT 1

   UNION ALL

   SELECT resources.*, count(clicks.id) as click_counts, 2 as SortKey
   FROM resources 
   INNER JOIN clicks ON clicks.clickable_id = resources.id AND clicks.clickable_type='Resource'
   WHERE (resources.mediable_id IN(1) AND resources.mediable_type='Level1') OR (resources.mediable_id IN(1, 2, 3, 4, 5) AND resources.mediable_type='Level2')
   GROUP BY resources.id
   ORDER BY SortKey, mediable_type, click_counts) as t"

select_sql = "SELECT DISTINCT resources.*, click_counts FROM #{sub_select} LIMIT 50"

results = ActiveRecord::Base.connection.select_all(select_sql).rows # array
results_ordered = results.sort { |a, b| a.last <=> b.last } # sort by clicks_count

注意事项：

• SortKey 额外属性保证查询顺序
• UNION ALL 不会消除任何重复记录。所以我们需要在resources.*, click_counts 列上添加DISTINCT 以删除可能的重复记录（'Level2' 的第一个）

希望对你有帮助！