获取REST资源作为列表< T>与泽西岛 [英] Fetching REST resource as List<T> with Jersey

问题描述

我试图在Jersey中编写一个通用函数，它可以用来通过REST获取相同类型的对象列表。我将其基于本论坛中的信息：链接

  @Override 
 public< T>列表与LT; T> fetchResourceAsList（String url）{
 ClientConfig cc = new DefaultClientConfig（）; 
客户端c = Client.create（cc）; 
 if（userName！= null&& password！= null）{
 c.addFilter（new HTTPBasicAuthFilter（userName，password））; 
} 
 WebResource resource = c.resource（url）; 
 return resource.get（new GenericType< List< T>>（）{}）; 
} 
  

然而这不起作用。如果我尝试执行它，我会得到以下错误： SEVERE：Java类java.util.List和Java类型java.util.List< T>和MIME媒体类型的消息正文阅读器没有找到application / xml 。

然而，如果我没有模板编写这个函数（用一个实际的类名替换T），它可以正常工作。当然，这种方式失去了它的意义。

有没有办法解决这个问题？

解决方案

我找到了解决方案
https://java.net/projects/jersey/lists/users/archive/2011-08/message/37

  public< T>列表与LT; T> getAll（final Class< T> clazz）{
 
 ParameterizedType parameterizedGenericType = new ParameterizedType（）{
 public Type [] getActualTypeArguments（）{
 return new Type [] {clazz} ; 
} 
 
 public Type getRawType（）{
 return List.class; 
} 
 
 public Type getOwnerType（）{
 return List.class; 
} 
}; 
 
 GenericType< List< T>> genericType = new GenericType< List< T>>（
 parameterizedGenericType）{
}; 
 
 return service.path（Path.ROOT）.path（clazz.getSimpleName（））
 .accept（MediaType.APPLICATION_XML）.get（genericType）; 
} 
  

I'm trying to write a generic function in Jersey which can be used to fetch a List of objects of the same type through REST. I based it on the informations found in this forum: link

@Override
public <T> List<T> fetchResourceAsList(String url) {
  ClientConfig cc = new DefaultClientConfig();
  Client c = Client.create(cc);
  if (userName!=null && password!=null) {
    c.addFilter(new HTTPBasicAuthFilter(userName, password)); 
  }
  WebResource resource = c.resource(url);
  return resource.get(new GenericType<List<T>>() {});
}

However this is not working. If i try to execute it, i get the following error: SEVERE: A message body reader for Java class java.util.List, and Java type java.util.List<T>, and MIME media type application/xml was not found.

However if i write this function without templating (replacing T with an actual class name) it just works fine. Of course this way the function loses it's meaning.

Is there a way to fix this?

解决方案

I've found solution https://java.net/projects/jersey/lists/users/archive/2011-08/message/37

public <T> List<T> getAll(final Class<T> clazz) {

    ParameterizedType parameterizedGenericType = new ParameterizedType() {
        public Type[] getActualTypeArguments() {
            return new Type[] { clazz };
        }

        public Type getRawType() {
            return List.class;
        }

        public Type getOwnerType() {
            return List.class;
        }
    };

    GenericType<List<T>> genericType = new GenericType<List<T>>(
            parameterizedGenericType) {
    };

    return service.path(Path.ROOT).path(clazz.getSimpleName())
            .accept(MediaType.APPLICATION_XML).get(genericType);
}

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