红宝石在rails-使用范围内的联盟

Question

我使用rails 3.0.3。 我有4张桌子。

class Artwork 
    has_and_belongs_to_many :tag_styles 
    has_and_belongs_to_many :tag_subjects 
    has_and_belongs_to_many :tag_arttypes 
end 

class TagArttype 
    set_table_name :tag_arttypes 
    has_and_belongs_to_many :artworks 
end 

class TagStyle 
    set_table_name :tag_styles 
    has_and_belongs_to_many :artworks 
end 

class TagSubject 
    set_table_name :tag_subjects 
    has_and_belongs_to_many :artworks 
end 

所有这三个tag_tables都有一个name属性。 我怎样才能使它接受一个单词，并用下面的SQL将其放在“抽象”的地方艺术品范围：

select a.* from artworks as a, tag_arttypes as ta, artworks_tag_arttypes as ata where a.id = ata.artwork_id and ta.id = ata.tag_arttype_id and ta.name = 'Abstract' union 
select a.* from artworks as a,tag_styles as ta,artworks_tag_styles as ata where a.id = ata.artwork_id and ta.id = ata.tag_style_id and ta.name = 'Abstract'union 
select a.* from artworks as a,tag_subjects as ta,artworks_tag_subjects as ata where a.id = ata.artwork_id and ta.id = ata.tag_subject_id and ta.name = 'Abstract'; 

谢谢！

Answer 1

它有点迟了，但这可能对某人有所帮助。最后，找到了导轨的做法。

scope :by_tags, lambda { |tag| 
    includes(:tag_styles,:tag_subjects,:tag_arttypes).where('tag_styles.name LIKE ? OR tag_arttypes.name LIKE ? OR tag_subjects.name LIKE ?',"%#{tag}","%#{tag}","%#{tag}"). 
select("DISTINCT artworks.*") 
} 

Answer 2

它很简单，如果我正确理解你的问题。像下面的东西会工作：

scope :filter, lambda {|name| 
" 
(select a.* from artworks as a, tag_arttypes as ta, artworks_tag_arttypes as ata where a.id = ata.artwork_id and ta.id = ata.tag_arttype_id and ta.name = '#{name}') union 
(select a.* from artworks as a,tag_styles as ta,artworks_tag_styles as ata where a.id = ata.artwork_id and ta.id = ata.tag_style_id and ta.name = '#{name}') union 
(select a.* from artworks as a,tag_subjects as ta,artworks_tag_subjects as ata where a.id = ata.artwork_id and ta.id = ata.tag_subject_id and ta.name = '#{name}') 
" 
}