JavaEE如何从类声明它的API路径

Question

使用：Java EE + JAX-RS（Apache Wink）+ WAS。

让说，我有REST API的类中声明你好，路径"/hello"

@Path("/hello") 
public class Hello{ 

    @GET 
    @Produces(MediaType.APPLICATION_JSON) 
    public Response sayHello() { 
     Map<String, String> myMap = new LinkedHashMap<String, String>(); 
     myMap.put("firstName", "To"); 
     myMap.put("lastName", "Kra"); 
     myMap.put("message", "Hello World!"); 
     Gson gson = new Gson(); 
     String json = gson.toJson(myMap);  
     return Response.status(200).entity(json).build(); 
    } 
} 

我怎样才能从Hello.class这条道路不使用反射？我可以在javax.ws.rs.core.UriBuilder方法path(Class clazz)中看到例子，它可以以某种方式得到它，找不到它的来源。

Answer 1

解决方案与思考：

/** 
* Gets rest api path from its resource class 
* @param apiClazz 
* @return String rest api path 
*/ 
public static String getRestApiPath(Class<?> apiClazz){ 
    Annotation[] annotations = apiClazz.getAnnotations(); 
    for(Annotation annotation : annotations){ 
     if(annotation instanceof Path){ 
      Path pathAnnotation = (Path) annotation; 
      return pathAnnotation.value(); 
     } 
    } 
    return ""; 
} 

Answer 2

添加@Context要么方法调用或类和注射两种HttpServletRequest或UriInfo，无论将更加有用，就像这样：

// as class fields 
@Context 
private HttpServletRequest request; 

@Context 
private UriInfo uriInfo; 
... 

// or as param in method 
@GET 
@Produces(MediaType.APPLICATION_JSON) 
public Response sayHello(@Context UriInfo uriInfo) { 
.... 

System.out.println(request.getRequestURI()); 
System.out.println("uri: " + uriInfo.getPath()); 
System.out.println("uri: " + uriInfo.getBaseUri());