2013-07-29 36 views
0

我的應用程序詢問用戶的飼料URL地址。我試圖將Feed url傳遞給Feedzirra來獲取並解析url,但我沒有任何運氣。使用模型變量通過Feedzirra拉動飼料

我在做什麼錯:

這裏是我的節目從控制器方法

# GET /scores/1 
# GET /scores/1.json 
def show 

#@score=Score.find(params[:id]) 
@score = current_user.scores.find(params[:id]) 

url=:url 
feed=Feedzirra::Feed.fetch_and_parse(url) 
@entry=feed.entries 

respond_to do |format| 
    format.html #show.html.erb 
    format.json {render json: @score } 
end 

型號

class Score < ActiveRecord::Base 
    belongs_to :user 
    validates :user_id, presence: true 
    attr_accessible :title, :postFrequency, :editorialCalendar, :postTemplate, 
    :headlineVariations, :publishingScore, :professionalLogo, :colorScheme, :typographyStandards, 
    :seoPlugin, :googleAnalytics, :setupScore, :sb_promotion_1, :sb_promotion_2, :sb_recentPosts, 
    :sidebarScore, :leadMagnet, :emailCampaign, :productCreated, :landingPage, :monetizationScore, :url, 
    :name, :summary, :url, :published_at, :guid 

end 

回答

0
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這裏url變量總是會用一個符號的值:url。不要以爲這是你想要的。