Question

我正在尝试通过REST Web服务传递对象。以下是我的课程使用一些示例代码解释了我需要的功能。

Rest Web Service Class方法

@POST
@Path("/find")
@Consumes(MediaType.APPLICATION_FORM_URLENCODED)
@Produces({MediaType.APPLICATION_JSON})
public Response getDepartments(){
    Response response = new Response();

    try {

        response.setCode(MessageCode.SUCCESS);
        response.setMessage("Department Names");
        Department dept = new Department("12", "Financial");
        response.setPayload(dept);

    } catch (Exception e) {
        response.setCode(MessageCode.ERROR);
        response.setMessage(e.getMessage());
        e.printStackTrace();
    }       
    return response;
}

响应级

import java.io.Serializable;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement

public class Response implements Serializable{

    private static final long serialVersionUID = 1L;

    public enum MessageCode {
        SUCCESS, ERROR, UNKNOWN
    }

    private MessageCode code;
    private String message;
    private Object payload;

    public MessageCode getCode() {
        return code;
    }

    public void setCode(MessageCode code) {
        this.code = code;
    }

    public String getMessage() {
        return message;
    }

    public void setMessage(String message) {
        this.message = message;
    }

    public Object getPayload() {
        return payload;
    }

    public void setPayload(Object payload) {
        this.payload = payload;
    }
}

系类

@XmlRootElement
public class Department implements java.io.Serializable {


    private String deptNo;
    private String deptName;


    public Department() {
    }

    public Department(String deptNo, String deptName) {
        this.deptNo = deptNo;
        this.deptName = deptName;
    }

    public String getDeptNo() {
        return this.deptNo;
    }

    public void setDeptNo(String deptNo) {
        this.deptNo = deptNo;
    }

    public String getDeptName() {
        return this.deptName;
    }

    public void setDeptName(String deptName) {
        this.deptName = deptName;
    }

}

当我在其余的Web服务类中调用getDepartments方法时，它返回以下异常。但是，如果我在Response类中将有效负载的类型对象更改为部门，则会正确返回json响应。但是因为我需要将此Response类用于不同类型的类，所以我无法将有效负载重新绑定为一个类类型。有人可以帮我解决这个问题吗？

堆栈跟踪

Dec 27, 2012 9:34:18 PM com.sun.jersey.spi.container.ContainerResponse logException
SEVERE: Mapped exception to response: 500 (Internal Server Error)
javax.ws.rs.WebApplicationException: javax.xml.bind.MarshalException
 - with linked exception:
[javax.xml.bind.JAXBException: class Department nor any of its super class is known to this context.]
    at com.sun.jersey.core.provider.jaxb.AbstractRootElementProvider.writeTo(AbstractRootElementProvider.java:159)
    at com.sun.jersey.spi.container.ContainerResponse.write(ContainerResponse.java:306)
    at com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1437)
    at com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1349)
    at com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1339)
    at com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:416)
    at com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:537)
    at com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:699)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:820)
    at org.mortbay.jetty.servlet.ServletHolder.handle(ServletHolder.java:511)
    at org.mortbay.jetty.servlet.ServletHandler.handle(ServletHandler.java:401)
    at org.mortbay.jetty.security.SecurityHandler.handle(SecurityHandler.java:216)
    at org.mortbay.jetty.servlet.SessionHandler.handle(SessionHandler.java:182)
    at org.mortbay.jetty.handler.ContextHandler.handle(ContextHandler.java:766)
    at org.mortbay.jetty.webapp.WebAppContext.handle(WebAppContext.java:450)
    at org.mortbay.jetty.handler.HandlerWrapper.handle(HandlerWrapper.java:152)
    at org.mortbay.jetty.Server.handle(Server.java:326)
    at org.mortbay.jetty.HttpConnection.handleRequest(HttpConnection.java:542)
    at org.mortbay.jetty.HttpConnection$RequestHandler.content(HttpConnection.java:945)
    at org.mortbay.jetty.HttpParser.parseNext(HttpParser.java:756)
    at org.mortbay.jetty.HttpParser.parseAvailable(HttpParser.java:218)
    at org.mortbay.jetty.HttpConnection.handle(HttpConnection.java:404)
    at org.mortbay.io.nio.SelectChannelEndPoint.run(SelectChannelEndPoint.java:410)
    at org.mortbay.thread.QueuedThreadPool$PoolThread.run(QueuedThreadPool.java:582)
Caused by: javax.xml.bind.MarshalException
 - with linked exception:
[javax.xml.bind.JAXBException: class Department nor any of its super class is known to this context.]
    at com.sun.xml.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:323)
    at com.sun.xml.bind.v2.runtime.MarshallerImpl.marshal(MarshallerImpl.java:177)
    at com.sun.jersey.json.impl.BaseJSONMarshaller.marshallToJSON(BaseJSONMarshaller.java:103)
    at com.sun.jersey.json.impl.provider.entity.JSONRootElementProvider.writeTo(JSONRootElementProvider.java:136)
    at com.sun.jersey.core.provider.jaxb.AbstractRootElementProvider.writeTo(AbstractRootElementProvider.java:157)
    ... 23 more
Caused by: javax.xml.bind.JAXBException: class Department nor any of its super class is known to this context.
    at com.sun.xml.bind.v2.runtime.XMLSerializer.reportError(XMLSerializer.java:250)
    at com.sun.xml.bind.v2.runtime.XMLSerializer.reportError(XMLSerializer.java:265)
    at com.sun.xml.bind.v2.runtime.XMLSerializer.childAsXsiType(XMLSerializer.java:657)
    at com.sun.xml.bind.v2.runtime.property.SingleElementNodeProperty.serializeBody(SingleElementNodeProperty.java:156)
    at com.sun.xml.bind.v2.runtime.ClassBeanInfoImpl.serializeBody(ClassBeanInfoImpl.java:344)
    at com.sun.xml.bind.v2.runtime.XMLSerializer.childAsSoleContent(XMLSerializer.java:597)
    at com.sun.xml.bind.v2.runtime.ClassBeanInfoImpl.serializeRoot(ClassBeanInfoImpl.java:328)
    at com.sun.xml.bind.v2.runtime.XMLSerializer.childAsRoot(XMLSerializer.java:498)
    at com.sun.xml.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:320)
    ... 27 more
Caused by: javax.xml.bind.JAXBException: class Department nor any of its super class is known to this context.
    at com.sun.xml.bind.v2.runtime.JAXBContextImpl.getBeanInfo(JAXBContextImpl.java:611)
    at com.sun.xml.bind.v2.runtime.XMLSerializer.childAsXsiType(XMLSerializer.java:652)
    ... 33 more

Answer 1

JAX-RS实现自动支持基于可发现的JAXB注释的类的编组/解组，但由于您的有效负载被声明为Object，我认为创建的JAXBContext错过了Department类什么时候发动它不知道怎么做。

快速而肮脏的修复方法是在响应类中添加XmlSeeAlso注释：

@XmlRootElement
@XmlSeeAlso({Department.class})
public class Response implements Serializable {
  ....

或者更复杂的东西是“通过Response来丰富”ContextResolver类的JAXB上下文：

import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.ext.ContextResolver;
import javax.ws.rs.ext.Provider;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;

@Provider
@Produces({ MediaType.APPLICATION_JSON, MediaType.APPLICATION_XML })
public class ResponseResolver implements ContextResolver<JAXBContext> {
    private JAXBContext ctx;

    public ResponseResolver() {
        try {
            this.ctx = JAXBContext.newInstance(

                        Response.class, 
                        Department.class

                    );
        } catch (JAXBException ex) {
            throw new RuntimeException(ex);
        }
    }

    public JAXBContext getContext(Class<?> type) {
        return (type.equals(Response.class) ? ctx : null);
    }
}

Answer 2

我有同样的问题，我通过添加包来探索Jaxb2marshaller来解决它。对于spring，将定义一个这样的bean：

@Bean
    public Jaxb2Marshaller marshaller() {
        Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
        String[] packagesToScan= {"<packcge which contain the department class>"};
        marshaller.setPackagesToScan(packagesToScan);
        return marshaller;
    }

通过这种方式，如果所有请求和响应类都在同一个包中，则不需要在JAXBcontext上专门指出类

Answer 3

可以通过指定完整的类路径来解决此异常。

示例：

如果您使用的是名为ExceptionDetails
的班级

传递参数的方法错误

JAXBContext jaxbContext = JAXBContext.newInstance(ExceptionDetails.class);

传递参数的正确方法

JAXBContext jaxbContext = JAXBContext.newInstance(com.tibco.schemas.exception.ExceptionDetails.class);

Answer 4

我在使用JAXB参考实现和JBoss AS 7.1时遇到了类似的问题。我能够编写一个集成测试，确认JAXB在JBoss环境之外工作（表明问题可能是JBoss中的类加载器）。

这是给出错误的代码（即无效）：

private static final JAXBContext JC;

static {
    try {
        JC = JAXBContext.newInstance("org.foo.bar");
    } catch (Exception exp) {
        throw new RuntimeException(exp);
    }
}

这是有效的代码（ValueSet是从我的XML封送的类之一）。

private static final JAXBContext JC;

static {
    try {
        ClassLoader classLoader = ValueSet.class.getClassLoader();
        JC = JAXBContext.newInstance("org.foo.bar", classLoader);
    } catch (Exception exp) {
        throw new RuntimeException(exp);
    }
}

在某些情况下，我获得了Class，也没有任何超类在这种情况下已知。在其他情况下，我也得到了org.foo.bar.ValueSet的异常，无法强制转换为org.foo.bar.ValueSet（类似于此处描述的问题：ClassCastException when casting to the same class）。

Answer 5

在支持一个应用程序时，我遇到了类似的错误。这与为SOAP Web服务生成的类有关。

此问题是由于缺少类引起的。当javax.xml.bind.Marshaller试图封送jaxb对象时，它没有找到由wsdl和xsd生成的所有依赖类。在将class包含所有类的jar添加到类路径后，问题得以解决。

Answer 6

就像@ftrujillo 已经提出的那样，如果您使用 Spring Jaxb2Marshaller，您可以设置要扫描的包。在某些情况下，如果有更复杂的带有继承的 XSD 模型或者只是具有更广泛传播的模型，那么在类层次结构中获取要扫描的包的信息可能会出现问题，那么您可以使用对象层次结构的扫描.在这种情况下，我利用了以下功能，处理简单的 POJO 层次结构（我相信它可能会针对非简单 POJO 进行增强）：

....
            Jaxb2Marshaller locMarshaller = new Jaxb2Marshaller();
            locMarshaller.setPackagesToScan(
                    resolveJaxb2MarshallerPackagesToScan(
                            someFaultContentContainerClass,
                            someFaultContentClass
                    )
            );
...

    /**
     * Resolves the packages which have to be passed into the Spring Jaxb2Marshaller
     * to be able to perform the serialization/deserialization action
     * @param aClassesToBeMarshalled classes which has to be marshalled. Null array items
     *   are ignored without any error.
     * @return packages which have to be scanned by Spring Jaxb2Marshaller for it.
     */
    public static String[] resolveJaxb2MarshallerPackagesToScan(Class<?>... aClassesToBeMarshalled) {
        List<String> locResultList = new ArrayList<>();
        for (Class<?> locProcessedTopLevelClass : aClassesToBeMarshalled) {
            Class<?> locProcessedClass = locProcessedTopLevelClass;
            while (locProcessedClass != null) {
                if (!locResultList.contains(locProcessedClass.getPackageName()))
                    locResultList.add(locProcessedClass.getPackageName());
                locProcessedClass = locProcessedClass.getSuperclass();
            }
        }
        return locResultList.toArray(new String[0]);
    }

Answer 7

Ftrujillo's answer效果很好，但是如果您只有一个要扫描的软件包，这是最短的形式：

    @Bean
    public Jaxb2Marshaller marshaller() {
        Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
        marshaller.setContextPath("your.package.to.scan");
        return marshaller;
    }

Answer 8

我知道这是一个老问题，但是您可以使用参数（P）更改响应：

public class Response<P> implements Serializable{

private static final long serialVersionUID = 1L;

public enum MessageCode {
    SUCCESS, ERROR, UNKNOWN
}

private MessageCode code;
private String message;
private P payload;

...
public P getPayload() {
    return payload;
}

public void setPayload(P payload) {
    this.payload = payload;
}

}

方法应该是

public Response<Departments> getDepartments(){...}

我现在无法尝试，但应该可以。

否则，可以扩展响应

@XmlRootElement    
public class DepResponse extends Response<Department> {<no content>}

Answer 9

当我们对Jaxb2Marshaller使用相同的方法名称时，会发生此错误例如：

    @Bean
    public Jaxb2Marshaller marshallerClient() {
        Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
        // this package must match the package in the <generatePackage> specified in
        // pom.xml
        marshaller.setContextPath("library.io.github.walterwhites.loans");

        return marshaller;
    }

在其他文件上

    @Bean
    public Jaxb2Marshaller marshallerClient() {
        Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
        // this package must match the package in the <generatePackage> specified in
        // pom.xml
        marshaller.setContextPath("library.io.github.walterwhites.client");

        return marshaller;
    }

即使是不同的类别，您也应该使用不同的名称

javax.xml.bind.JAXBException：此类上下文中已知类***或其任何超类

9 个答案: