标签: jax-rs

我正在使用Jersey的集成Jackson处理将传入的JSON转换为POJO,例如:

@POST
@Consumes(MediaType.APPLICATION_JSON)
public Response newCustomer( CustomerRepresentation customer)
{
...
}

如果客户端发送带有无效字段的JSON,则Jersey当前返回a 500 Internal Server Error.相反,我想返回一个400 Bad Request,最好有一些有意义的细节,表明哪些字段有误.

有关如何实现这一目标的任何见解？(至少返回一个通用的400而不是完全不合适的500？)

更新: 这是在调用我的处理程序之前在服务器端生成的异常:

javax.servlet.ServletException: org.codehaus.jackson.map.exc.UnrecognizedPropertyException: 
Unrecognized field "this_isnt_a_known"_field" (Class com.redacted....), not marked as ignorable

非常直截了当的问题.我正在使用Jersey构建一个REST系统.如果我有一个具有值的类,我需要在处理期间使用但不希望在类被封送时作为XML或JSON输出的一部分发送,有没有办法忽略它？就像是:

@XmlRootElement(name="example")
class Example {
    private int a;
    private String b;
    private Object c;

    @XmlElement(ignore=true)
    public int getA() { return a; }
    @XmlElement
    public String getB() { return b; }
    @Ignore
    public Object getC() { return c; }
    ... //setters, constructors, etc.
}

我希望像ignore=trueover getA()或@Ignoreover 这样的东西getC()能起作用,但我找不到任何文档.

从Jersey 2.9开始,可以通过声明性链接为超媒体驱动的REST API创建链接关系.

这段代码,例如:

@InjectLink(
    resource = ItemResource.class,
    style = Style.ABSOLUTE,
    bindings = @Binding(name = "id", value = "${instance.id}"),
    rel = "self"
)
@XmlJavaTypeAdapter(Link.JaxbAdapter.class)
@XmlElement(name="link")
Link self;

......理论上预计会产生这样的JSON:

"link" : {
    "rel" : "self",
    "href" : "http://localhost/api/resource/1"
}

但是,Jersey会生成不同的JSON,其中包含许多我不需要的属性:

"link" : {
   "rel" : "self",
   "uri" : "http://localhost/api/resource/1",
   "type": null,
   "uriBuilder" : null
}

另请注意href,它使用的是代替uri.我查看了Jersey Link对象的实现并找到了JerseyLink.

我想使用Jersey的声明性链接,而不是推出我自己的实现.我最终使用Jackson注释只是为了忽略其他JerseyLink属性.

@JsonIgnoreProperties({ "uriBuilder", "params", "type", "rels" })

有没有人使用与Jersey的声明性链接并且具有预期的JSON输出(例如,href而不是uri没有额外的Jersey属性)而不必使用JsonIgnoreProperties或其他黑客？

谢谢. …

我正在使用JAX-RS我的网络服务.我有共同的功能,并希望使用继承.我提供简单的CRUD操作.我已经定义了这样的接口:

public interface ICRUD {

    @POST
    @Consumes("application/json")
    @Produces("application/json")
    @Path("create")
    public String createREST(String transferObject);

    @GET
    @Consumes("application/json")
    @Produces("application/json")
    @Path("retrieve/{id}")
    public String retrieveREST(@PathParam("id") String id);

    @POST
    @Consumes("application/json")
    @Produces("application/json")
    @Path("update")
    public void updateREST(@Suspended final AsyncResponse asyncResponse,
                           final String transferObject) ;

    @DELETE
    @Consumes("application/json")
    @Produces("application/json")
    @Path("delete/{id}")
    public String deleteREST(@PathParam("id") String id); 
}

我有一个实现此接口的抽象类:

public abstract class BaseREST implements ICRUD{

private final ExecutorService executorService = Executors.newCachedThreadPool();

@Override
public String createREST(String transferObject) {
    return create(transferObject).toJson();
}

@Override
public String retreiveREST(@PathParam("id") String id) {
    return retreive(id).toJson();
} …

我在下面给出了一个POJO,我希望将其作为JSON或XML输出到服务器.

这就是我所做的

客户:

ClientConfig config = new ClientConfig();
Client client = ClientBuilder.newClient(config);
WebTarget target = client.target(getBaseURI());

public void putFriend(String uri , Friend friend)
{
    System.out.println(friend.toString());

    target = target.path(some_path).path(uri);
    ClientResponse response =        target.request(MediaType.APPLICATION_JSON).put(Entity.entity(friend,MediaType.APPLICATION_JSON),ClientResponse.class);
}

我在网上找到的例子是使用WebResource.

我不知道如何使用WebTarget.我所做的是从在SO上找到的一些示例中获取,但Entity.entity()给出了错误未定义的方法实体(friend,String).

POJO

@XmlRootElement
public class Friend{

    private String friendURI;
    private String event;
    private String uri;

    String getUri() {
        return uri;
    }
    void setUri(String uri) {
        this.uri = uri;
    }
    String getFriendURI() {
        return friendURI;
    }
    void setFriendURI(String friendURI) {
        this.friendURI = friendURI;
    }
    String getEvent() { …

我试图了解DropWizard中的身份验证和授权是如何工作的.我已经阅读了他们的auth指南以及GitHub上的dropwizard-security项目,但感觉我仍然缺少一些重要的概念.

public class SimpleCredential {
    private String password;

    public SimpleCredential(String password) {
        super();

        this.password = password;
    }
}

public class SimplePrincipal {
    pivate String username;

    public SimplePrincipal(String username) {
        super();

        this.username = username;
    }
}

public class SimpleAuthenticator implements Authenticator<SimpleCredential, SimplePrincipal> {
    @Override
    public Optional<SimplePrincipal> authenticate(SimpleCredential credential) throws AuthenticationException {
        if(!"12345".equals(credential.getPassword())) {
            throw new AuthenticationException("Sign in failed.");
        }

        Optional.fromNullable(new SimplePrincipal("simple_user"));
    }
}

然后在我的Application子类中:

@Override
public void run(BackendConfiguration configuration, Environment environment) throws Exception …

我正在使用带有jackson提供程序的RestEasy客户端并收到上述错误

客户端代码是:

ClientRequest request = new ClientRequest(url);
request.accept(MediaType.APPLICATION_JSON);
ClientResponse<String> response = request.get(String.class);

if (response.getStatus() != 200) {
  throw new RuntimeException("Failed : HTTP error code : " + response.getStatus());
}

BufferedReader br =
  new BufferedReader(new InputStreamReader(new ByteArrayInputStream(response.getEntity().getBytes())));

response.getEntity()正在抛出ClientResponseFailure异常,错误是

Unable to find a MessageBodyReader of content-type application/json and type class java.lang.String

我的服务器端代码如下:

@GET
@Path("/{itemId}")
@Produces(MediaType.APPLICATION_JSON)
public String item(@PathParam("itemId") String itemId) {
  //custom code

  return gson.toJSON(object);
}

我正在尝试使用Jersey Client来模拟对我的Web服务的HTTP请求.我试图从文档中实现这个简单的例子.这是我的简短代码:

public void restoreTest(String sessionId) throws Exception {
    Client client = ClientBuilder.newClient();
    WebTarget target = client.target(idsUrl).path("restore");
    Form form = new Form();
    form.param("sessionId", sessionId);
    target.request(MediaType.APPLICATION_FORM_URLENCODED_TYPE);
}

我甚至没有实现整个示例,因为目前我在最后一行得到一个异常:

java.lang.NoSuchMethodError: javax.ws.rs.core.MultivaluedMap.addAll(Ljava/lang/Object;[Ljava/lang/Object;)V
    at org.glassfish.jersey.client.ClientRequest.accept(ClientRequest.java:254)
    at org.glassfish.jersey.client.JerseyWebTarget.request(JerseyWebTarget.java:232)
    at org.glassfish.jersey.client.JerseyWebTarget.request(JerseyWebTarget.java:60)
    at org.icatproject.idsclient.TestingClient.restoreTest(TestingClient.java:112)
    at org.icatproject.ids.ids2.ArchiveTest.restoreThenArchiveDataset(ArchiveTest.java:55)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:616)
    at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:44)
    at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:15)
    at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:41)
    at org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:20)
    at org.junit.internal.runners.statements.RunBefores.evaluate(RunBefores.java:28)
    at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:76)
    at org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:50)
    at org.junit.runners.ParentRunner$3.run(ParentRunner.java:193)
    at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:52)
    at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:191)
    at org.junit.runners.ParentRunner.access$000(ParentRunner.java:42)
    at org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:184)
    at org.junit.internal.runners.statements.RunBefores.evaluate(RunBefores.java:28)
    at org.junit.runners.ParentRunner.run(ParentRunner.java:236)
    at org.eclipse.jdt.internal.junit4.runner.JUnit4TestReference.run(JUnit4TestReference.java:50) …

如何Map使用Jersey/JAX-RS框架将XML作为XML/JSON文档返回并不是那么明显.它已经支持Lists,但是当涉及到Maps时,则没有MessageBodyWriter.即使我要将其嵌入Ma到包装类中,mapXML模式中也没有类型.

关于如何将Map编组到Jersey中的XML/JSON文档的任何实用建议？

我想在球衣休息服务中抓住所有意想不到的例外情况.因此我写了一个ExceptionMapper:

@Provider
public class ExceptionMapper implements javax.ws.rs.ext.ExceptionMapper<Exception> {
    private static Logger logger = LogManager.getLogManager().getLogger(ExceptionMapper.class.getName());

    @Override
    public Response toResponse(Exception e) {
        logger.log(Level.SEVERE, e.getMessage(), e);

        return Response.status(Response.Status.INTERNAL_SERVER_ERROR).entity("Internal error").type("text/plain").build();
    }
}

映射器捕获所有异常.所以我写不出来:

public MyResult getById(@PathParam("id")) {
    if (checkAnyThing) {
        return new MyResult();
    }
    else {
        throw new WebApplicationException(Response.Status.NOT_FOUND);
    }
}

这是由Mapper捕获的.现在我要写:

public Response getById(@PathParam("id") {
    if (checkAnyThing) { {
        return Response.ok().entity(new MyResult()).build();
    }
    else {
        return Response.status(Response.Status.NOT_FOUND).build();
    }
}

这是捕获所有意外异常并正确返回错误(错误代码)的正确方法吗？或者还有其他(更正确的)方式吗？