在 Rails 3 中使用多个可选的 where 条件(子句)进行查询

Question

如何按多个参数进行搜索/过滤，这些参数在 Rails 3.2 中都是可选的？ 通过下面的设置，我目前收到以下错误。 任何帮助将不胜感激。

undefined method `paginate' for nil:NilClass
Application Trace | Framework Trace | Full Trace
app/controllers/contacts_controller.rb:50:in `index'

这是我contacts_controller中的索引操作：

def index
  city = params[:city]
  state = params[:state]
  zip = params[:zip]
  @contacts = Contact.search(city,state,zip).paginate(:page => params[:page], :per_page => items_per_page)
end

这是我的联系模型中的搜索方法：

def self.search(city, state, zip)
  joins(:profile => :addresses)
  .where("city like ?", "%#{city}%") unless city.blank?
  .where("state = ?", state) unless state.blank?
  .where("zip like ?", "%#{zip}%") unless zip.blank?
end

Answer 1

尝试更明确地返回，例如：

def self.search(city, state, zip)
  query_obj = joins(:profile => :addresses)
  query_obj = query_obj.where("city like ?", "%#{city}%") unless city.blank?
  query_obj = query_obj.where("state = ?", state) unless state.blank?
  query_obj = query_obj.where("zip like ?", "%#{zip}%") unless zip.blank?
  query_obj
end

Answer 2

def self.search(city, state, zip)
  conditions = {:"city like %?%" => city, :state => state, :"zip like %?%" => zip}
  conditions.delete_if {|key,val| val.blank? }
  self.joins(:profile => :addresses)
  .where(conditions)
end

和你的函数 Contact.search（city，state，zip） 返回 nil，这就是为什么它给出 nil.paginate 的错误。