Question

我有这样的模型，具有多态关系

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，我会在某处显示这些媒体，用户可以点击此处，每次点击我都会在点击表中保存一个条目

现在我需要在level1的显示页面上的用户时，我需要根据点击次数显示level1s和level2s的前50个资源，并且至少有一个资源将从数据库中提取

我将尝试这样：

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，但不能保证我至少有一个与级别2相关的资源。

你能帮我解决这个问题吗？

有可能从未点击资源，但我必须获得资源以及每个级别至少需要一个资源所以我认为内部联接应该改为左外部

Answer 1

如果您不需要进行额外的计算，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，您可以尝试这样的事情（未经测试）：

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：

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

希望它有所帮助！

如何获取大多数点击的记录和至少一个子级资源将包含

1 个答案: