[英]Issue with Rails link_to if method
在我的Ruby on Rails應用程序中,我具有如下所示的方法:
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
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
我試圖對其進行重構,並使用Rails 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
但是此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]
為什么此代碼不起作用?
嘗試顯式設置id
:
admin_survey_pack_sign_off_path(id: sp.survey_pack_sign_off)
如果我正確理解您的代碼,並且survey_pack_sign_off
包含頁面的某些ID
解決方案是:
spso = sp.survey_pack_sign_off
link_to_if spso, signed_off, spso ? admin_survey_pack_sign_off_path(spso) : nil
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