java - javax.xml.bind.JAXBException: Class *** 或其任何超类对此上下文都是已知的

Question

我正在尝试通过 REST Web 服务传递一个对象。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 服务类方法

@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;
}

响应类

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;
    }
}

部门班

@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 Web 服务类中调用 getDepartments 方法时，它会返回以下异常。但是如果我在响应类中将有效负载的类型对象更改为部门，它会正确返回 json 响应。但是由于我需要将这个 Response 类用于不同类型的类，所以我无法将有效负载重新字符串化为一种类类型。任何人都可以在这件事上帮助我吗？

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

score 75 · Accepted Answer

JAX-RS 实现自动支持基于可发现的 JAXB 注释的类的编组/解组，但是由于您的有效负载被声明为Object，我认为创建的类JAXBContext错过了Department类，并且何时编组它不知道如何。

一个快速而肮脏的修复方法是XmlSeeAlso向您的响应类添加注释：

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

或者更复杂一点的事情是Response通过使用 a 来“丰富”类的 JAXB 上下文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);
    }
}

score 17 · Accepted Answer

我有同样的问题，我通过向 Jaxb2marshaller 添加要探索的包来解决它。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，如果您的所有请求和响应类都在同一个包中，则不需要在 JAXBcontext 上特别指出类

score 5 · Accepted Answer

可以通过指定完整的类路径来解决此异常。

例子：

如果您使用的是名为ExceptionDetails

传递参数的错误方式

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

传递论据的正确方法

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

score 2 · Accepted Answer

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（暗示问题可能是 JBoss 中的类加载器）。

这是给出错误的代码（即不工作）：

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 编组的类之一）。

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，我得到了这个类，也没有它的任何超类在这种情况下是已知的。在其他情况下，我也遇到了 org.foo.bar.ValueSet cannot be cast to org.foo.bar.ValueSet 的异常（类似于此处描述的问题：ClassCastException when cast to the same class）。

score 2 · Accepted Answer

在支持一个应用程序时，我遇到了类似的错误。这是关于为 SOAP Web 服务生成的类。

该问题是由于缺少课程而引起的。当 javax.xml.bind.Marshaller 试图编组 jaxb 对象时，它没有找到使用 wsdl 和 xsd 生成的所有依赖类。在类路径中添加包含所有类的 jar 后，问题已解决。

score 2 · Accepted Answer

Ftrujillo 的回答效果很好，但如果你只有一个包裹要扫描，这是最短的形式::

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

score 2 · Accepted Answer

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

    @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;
    }

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;
    }

即使是不同的班级，您也应该以不同的方式命名它们

score 1 · Accepted Answer

我知道这是一个老问题，但您可以使用参数 (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;
}

}

该方法是

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

我现在不能尝试，但应该可以。

否则可以扩展 Response

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

score 1 · Accepted Answer

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，如果有一个更复杂的具有继承的 XSD 模型，或者只是具有更广泛的模型，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，我利用了以下函数来处理普通的 POJO 层次结构（我相信它可能会针对非简单的 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]);
    }

score 0 · Accepted Answer

我有一个类似的错误，同样的错误信息，但不同的原因。

在我的情况下，问题是一个不完整的 ObjectFactory 实现，它自动创建为空。在为我的主类编写 ObjectFactory 后，问题就解决了。

@XmlRegistry
public class ObjectFactory {

    /**
     * Create a new ObjectFactory that can be used to create new instances of schema derived classes for package: x
     * 
     */
    public ObjectFactory() {
    }

    public CartaPorte createCartaPorte() {
        return new CartaPorte();
    }
}

java - javax.xml.bind.JAXBException: Class *** 或其任何超类对此上下文都是已知的

10 回答 10

Related

Reference