1

例如这个查询：

SELECT `variants`.* 
  FROM `variants` INNER JOIN `variant_attributes` 
    ON variant_attributes.variant_id = variants.id 
 WHERE (variant_attributes.id IN ('2','5'))

和变体has_many variant_attributes

我真正想做的是找到哪个变体同时具有 ID = 2 和 5 的变体属性。这对 MySQL 可行吗？额外的问题，有没有一种快速的方法来使用 Ruby on Rails，也许使用 SearchLogic？

解决方案

感谢 Quassnoi 提供的查询，效果很好。

在 Rails 上使用，我使用了下面的 named_scope，我认为这对于初学者来说更容易理解。

基本上，named_scope 将返回 {:from => x, :conditions => y} 并且上面的行用于设置 y 变量。

  named_scope :with_variant_attribute_values, lambda { |values|
    conditions = ["(
            SELECT  COUNT(*)
            FROM    `variant_attributes`
            WHERE   variant_attributes.variant_id = variants.id
                    AND variant_attributes.value IN (#{values.collect { |value| "?" }.join ", "})
            ) = ?
    "]
    conditions = conditions + values + [values.length]
    {
    :from => 'variants', 
    :conditions => conditions
  }}
4

4 回答 4

3

假设这variant_attributes (variant_id, id)是独一无二的：

SELECT  `variants`.*
FROM    `variants`
WHERE   (
        SELECT  COUNT(*)
        FROM    `variant_attributes`
        WHERE   variant_attributes.variant_id = variants.id
                AND variant_attributes.id IN ('2','5')
        ) = 2
于 2010-04-30T17:05:47.380 回答
1

Quassnoi 发布了 mysql 查询，可以满足您的需求。这是一个Variant模型的方法，它将做等效。我正在做两种方法，一种variant_attributes (variant_id, id)是唯一的组合，另一种是不是

独特的：

class Variant < ActiveRecord::Base
  has_many :variant_attributes
  named_scope :with_variant_attributes, lamda { |*ids|
     ids = ids.flatten
     if(ids.length>0)
       result = {:include => :variant_attributes}
       sql_params = {:length => ids.length,:ids => ids}
       result[:conditions] = ["(:length = (select count(*) from variant_attributes
                                          where id in (:ids))",sql_params]
       result
     else
       nil
     end
   }
end

非唯一

class Variant < ActiveRecord::Base
  has_many :variant_attributes
  named_scope :with_variant_attributes, lamda { |*ids|
     ids = ids.flatten
     if(ids.length>0)
       result = {:include => :variant_attributes}
       conditions = []
       sql_params = {}

       ids.each_with_index do |id,i|
         conditions << "( 1 = Select Count(*) from variant_attributes where id = :id#{i})"
         sql_params["id#{i}"] =  id
       end
       result[:conditions] = [ '(' + conditions.join(' AND ') + ')', sql_params]
       result
     else
       nil
     end
   }
end

可以通过以下方式使用：

# Returns all Variants with variant_attributes 1, 2, & 3
vars = Variant.with_variant_attributes(1,2,3) 

# Returns Variant 5 if it has attributes 3 & 5, or null if it doesn't
vars = Variant.with_variant_attributes(3,5).find_by_id(5)

#Returns Variants between 1 and 20 if that have an attribute of 2
vars = Variant.with_variant_attributes(2).find(:conditions => "id between 1 and 20")

#can accept a variable array of ids
my_ids = [3,5]
vars = Variant.with_variant_attributes(my_ids)

此代码尚未经过测试。

于 2010-04-30T19:06:58.647 回答
0

我会为此创建一个named_scope：

class Variant < ActiveRecord::Base
  has_many    :variant_attributes

  named_scope :with_variant_attributes, lambda { |*ids| { 
            :joins => :variant_attributes, 
            :conditions => {:variant_attributes=>{:id=>ids}},
            :group => "variants.id",
            :having => "count(variants.id) = #{ids.size}"
            }
          }
end

现在您可以按如下方式使用命名范围：

Variant.with_variant_attributes(1,2)
Variant.with_variant_attributes(1,2,3,4)
于 2010-04-30T22:13:05.363 回答
0

感谢 Quassnoi 提供的查询，效果很好。

在 Rails 上使用，我使用了下面的 named_scope，我认为这对于初学者来说更容易理解。

基本上，named_scope 将返回 {:from => x, :conditions => y} 并且上面的行用于设置 y 变量。

  named_scope :with_variant_attribute_values, lambda { |values|
    conditions = ["(
            SELECT  COUNT(*)
            FROM    `variant_attributes`
            WHERE   variant_attributes.variant_id = variants.id
                    AND variant_attributes.value IN (#{values.collect { |value| "?" }.join ", "})
            ) = ?
    "]
    conditions = conditions + values + [values.length]
    {
    :from => 'variants', 
    :conditions => conditions
  }}
于 2010-05-09T14:58:09.543 回答