Scala路由錯誤

Question

我是Scala和Play的新手。 我呼籲來自Application.scala功能在HTML頁面如下

<form action="@routes.Application.register()" class="pull-center"> 
     <div class="content"> 
      First Name:<input type="text" id="First Name"/> 
      Last Name: <input type="text" id="Last Name"/> 
      User Name: <input type="text" id="UserName"/> 
      Password: <input type="text" id="Password"/> 
      <button type="submit" class="btn">Register</button> 
     </div> 
</form> 

不過，我收到錯誤信息如下：

value register is not a member of controllers.ReverseApplication 

我指定的路線作爲如下路由文件如下：

# Routes 
# This file defines all application routes (Higher priority routes first) 
# ~~~~ 

# Home page 
GET /        controllers.Application.index 
GET  /room       controllers.Application.chatRoom(username: Option[String]) 
GET  /room/register   controllers.Application.register 
GET  /room/chat      controllers.Application.chat(username) 
GET  /assets/javascripts/chatroom.js controllers.Application.chatRoomJs(username: String) 

# Map static resources from the /public folder to the /assets URL path 
GET  /assets/*file     controllers.Assets.at(path="/public", file) 

在Application.scala該函數可以被定義如下：

def register() = Action { register => 
    Ok(views.html.register()) 

    } 

如何創建靜態URI？還有什麼是我的應用程序導致此錯誤出了什麼問題？

Answer 1

你可以給硬編碼的靜態URL

<form action="/room/register" class="pull-center"> 

但是當我想你的代碼，我也得到了同樣的錯誤，在月蝕，所以從清洗去除日食我的項目和賓果錯誤。
你可以忽略eclipse錯誤，你的代碼工作正常。