如何動態更改path_to（）？

Question

目前，我有我要崩潰了爲三種方式：

def send_email(contact,email) 

    end 

    def make_call(contact, call) 
    return link_to "Call", new_contact_call_path(:contact => contact, :call => call, :status => 'called') 
    end 

    def make_letter(contact, letter) 
    return link_to "Letter", new_contact_letter_path(:contact => contact, :letter => letter, :status => 'mailed') 
    end 

我想這三個摺疊成一個，這樣我就可以通過模型的參數之一，它仍然可以正確地創建了path_to。我想通過以下要做到這一點，但堅持：

def do_event(contact, call_or_email_or_letter) 
    model_name = call_or_email_or_letter.class.name.tableize.singularize 
    link_to "#{model_name.camelize}", new_contact_#{model_name}_path(contact, call_or_email_or_letter)" 
    end 

感謝的答案在這裏，我曾嘗試以下，這讓我更接近：

link_to("#{model_name.camelize}", send("new_contact_#{model_name}_path", 
             :contact => contact, 
             :status => "done", 
             :model_name => model_name)) 

但我似乎無法弄清楚如何在＃{model_name}爲「：」屬性時將它過去，然後將model_name的值發送，而不是字符串，而是引用該對象。

我得到這個工作： - 給點Kadada，因爲他在正確的方向:)

def do_event(contact, call_or_email_or_letter) 
    model_name = call_or_email_or_letter.class.name.tableize.singularize 
    link_to("#{model_name.camelize}", send("new_contact_#{model_name}_path", 
              :contact => contact, 
              :status => 'done', 
              :"#{model_name}" => call_or_email_or_letter))          
    end 

Answer 1

讓我試試這個：

def do_event(contact, call_or_email_or_letter) 
    model_name = call_or_email_or_letter.class.name.tableize.singularize 
    link_to("#{model_name.camelize}", send("new_contact_#{model_name}_path", 
       contact, call_or_email_or_letter)) 
end