問題和Rails的link_to如果方法

Question

在我的Ruby on Rails應用程序我有方法，看起來像這樣：

def survey_pack_signed_off(sp, type) 
    signed_off = Util::Boolean.humanize(sp.survey_pack_sign_off.present?)#returns Yes if true and No if false 
    return signed_off if type == :csv 
    if sp.survey_pack_sign_off.present? 
     link_to signed_off, admin_survey_pack_sign_off_path(sp.survey_pack_sign_off) 
    else 
    signed_off 
    end 
end 

我試圖重構它和使用Rails的link_to_if方法：

def survey_pack_signed_off(sp, type) 
    signed_off = Util::Boolean.humanize(sp.survey_pack_sign_off.present?)#returns Yes if true and No if false 
    return signed_off if type == :csv 
    link_to_if (type == :html && sp.survey_pack_sign_off.present?), signed_off, admin_survey_pack_sign_off_path(sp.survey_pack_sign_off) 
end 

但此link_to_if導致以下錯誤：

ActionController::UrlGenerationError: No route matches {:action=>"show", :controller=>"admin/survey_pack_sign_offs", :format=>nil, :id=>nil} missing required keys: [:id] 

爲什麼此代碼無效？

Answer 1

解決的辦法是：

spso = sp.survey_pack_sign_off 
    link_to_if spso, signed_off, spso ? admin_survey_pack_sign_off_path(spso) : nil 

Answer 2

嘗試設置id明確：

admin_survey_pack_sign_off_path(id: sp.survey_pack_sign_off) 

如果我理解正確的代碼，並survey_pack_sign_off包含頁面的一些ID