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?
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?
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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