mule-restlet可以將URL模式映射到特定方法嗎？

Question

要翻譯傳入的HTTP請求中Play! Web框架，具體方法是直截了當的，看起來像如下：

GET /users/{userId} UserController.getUser 
POST /users   UserController.addUser 
PUT /users   UserController.updateUser 
DELETE /users/{userId} UserController.deleteUser 

但我覺得很難做到這一點在騾子的Restlet。

<model name="userModel"> 
    <service name="userService"> 
     <inbound> 
      <inbound-endpoint address="http://localhost:63080"/> 
     </inbound> 
     <outbound> 
      <filtering-router> 
       <outbound-endpoint address="vm://userController"/> 
       <or> 
        <restlet:uri-template-filter pattern="https://stackoverflow.com/users/{userId}" verbs="GET"/> 
        <restlet:uri-template-filter pattern="/users" verbs="POST"/> 
        <restlet:uri-template-filter pattern="/users" verbs="PUT"/> 
        <restlet:uri-template-filter pattern="https://stackoverflow.com/users/{userId}" verbs="DELETE"/> 
       </or> 
      </filtering-router> 
     </outbound> 
    </service> 
    <service name="userController"> 
     <inbound> 
      <inbound-endpoint address="vm://userController"/> 
     </inbound> 
     <!-- **TODO: How to implement UserController** --> 
     <component class="com.ggd543.mulerestletdemo.user.UserController"/> 
    </service> 
</model> 

Answer 1

按照Restlet Transport doc，你應該能夠縮短你的路由映射到：

<or-filter> 
    <restlet:uri-template-filter pattern="/users" verbs="POST PUT"/> 
    <restlet:uri-template-filter pattern="https://stackoverflow.com/users/{userId}" verbs="GET DELETE"/> 
</or-filter> 

然後發展自己的UserController的資源，請參閱Restlet's user guide是相關的，由使用該版本運輸。