Question

这是原始逻辑

(scrape_datas = ScrapeData.find(
  :all, :conditions => 
  "artist_status = 'NOT_FOUND' 
   AND blacklisted = 1 
   AND extracted = 0 
   and not EXISTS(
     SELECT * FROM artist_name_suggestions where original = artist_name
   )

我能够更好地分割第一部分

scrape_datas = ScrapeData.where(
  :artist_status => 'NOT_FOUND',
  :blacklisted   => 1,
  :extracted     => 0
)

虽然在混合

中遇到“而不是EXISTS”查询时遇到问题

and not EXISTS(
  SELECT * FROM artist_name_suggestions where original = artist_name
)

谢谢！

Answer 1

首先，您可以提取简单的范围：

scope :not_found, where(:artist_status => 'NOT_FOUND')
scope :blacklisted, where(:blacklisted => 1)
scope :extracted, where(:extracted => 0)

然后添加一个查询方法（假设artist_name是一个scrape_datas列）：

def self.no_suggestions
  scrape_datas = ScrapeData.arel_table
  suggestions = ArtistNameSuggestion.arel_table
  where(ArtistNameSuggestion.where(
    suggestions[:original].eq(scrape_datas[:artist_name])
  ).exists.not)
end

现在你可以这样做：

ScrapeData.not_found.blacklisted.extracted.no_suggestions

Rails Arel相当于这个复杂的sql查询

1 个答案: