javax.xml.bind.JAXBException：***类或其任何超类中是众所周知的这种情况

我试图通过通过REST Web服务的对象。 以下是我的课解释说，我需要使用一些例如代码的功能。

REST Web服务类的方法

@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;
}

响应等级

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;
    }
}

科类

@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;
    }

}

当我做出getDepartments方法的调用在REST Web服务类返回下列情况除外。 但是，如果我改变有效载荷的类型对象 系 Response类是正确返回JSON响应。 但是，因为我需要使用这个Response类的不同类型的类的，我不能restring负荷到一个类的类型。 任何人都可以请帮我在这件事情？

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实现自动支持的基础上发现的JAXB注释类编组/解组，而是因为你的有效载荷被声明为Object ，我想创建JAXBContext错过了Department类，当它的时间马歇尔它不知道怎么办。

一个快速和肮脏的解决将是一个加XmlSeeAlso注释到您的响应类：

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

或更复杂的东西一点是“丰富”为JAXB的上下文Response通过使用一个类ContextResolver ：

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，我加入包探索到Jaxb2marshaller解决它。 因为春天将是这样定义一个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，如果所有的请求和响应类在同一个包，你并不需要特别指出的JAXBContext中的类

Answer 3:

此异常可以通过指定一个完整的类路径来解决。

例：

如果您使用的是类名为ExceptionDetails

---

传递参数错误方式

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

---

传递参数正确的方法

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

Answer 4:

我不得不使用JAXB参考实现和JBoss AS 7.1类似的问题。 我能够写一个集成测试证实JAXB工作的JBoss的环境之外（暗示问题可能是在JBoss的类加载器）。

这是一个被赋予错误（即不工作）的代码：

private static final JAXBContext JC;

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

这是工作的代码（ValueSet是从我的XML编组中的一个类）。

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，我得到了类或其任何超类是已知此背景下。 在其他情况下，我也得到了org.foo.bar.ValueSet的异常不能被转换为org.foo.bar.ValueSet（：类似的问题在这里描述的铸造同一类时抛出ClassCastException ）。

Answer 5:

支持一个应用程序时，我曾面对类似的错误。 这是关于生成的类为SOAP web服务。

这个问题是由于缺少类引起的。 当javax.xml.bind.Marshaller试图元帅JAXB对象它没有找到之后，通过WSDL和XSD所产生的所有相关类。 在类路径中的所有类添加JAR后，问题就解决了。

Answer 6:

Ftrujillo的回答运作良好，但如果你只有一个包进行扫描，这是最短的形式::

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

Answer 7:

我知道这是一个老问题，但你可以使用一个参数（P）变化的响应：

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;
}

}

该方法是

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

我现在不能尝试，但它应该工作。

否则，有可能延长响应

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

Answer 8:

当我们用同样的方法名称为Jaxb2Marshaller发生为例此错误：

    @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;
    }

和其他文件

    @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;
    }

即使这是不同的类，你应该将它们命名为不同

文章来源: javax.xml.bind.JAXBException: Class *** nor any of its super class is known to this context