jaxb - JAXB 编码对象 : javax. xml.bind.JAXBException : class . .. 或其任何父类(super class)的问题在此上下文中已知

Question

在我的 JAX-RS 项目( Jersey )中，我在将 JAXB 注释的对象之一编码为 JSON 时遇到问题。这是我在日志中看到的错误消息:

SEVERE: Internal server error javax.ws.rs.WebApplicationException: javax.xml.bind.JAXBException: class com.dnb.applications.webservice.mobile.view.CompaniesAndLocations nor any of its super class is known to this context.

这是否指向任何特定问题？我的资源有一个这样的方法:

@Path("/name/{companyname}/location/{location}")
@Produces("application/json; charset=UTF-8;")
@Consumes("application/json")
@POST
public Viewable findCompanyByCompanyNameAndLocationAsJSON(@PathParam("companyname") String companyName,
        @PathParam("location") String location, CriteriaView criteria) {
    criteria = criteria != null ? criteria : new CriteriaView();
    criteria.getKeywords().setCompanyName(companyName);
    return getCompanyListsHandler().listsByCompanyNameAndLocation(criteria, location);
}

Viewable是一个空界面。上述方法返回一个 CompaniesAndLocations 类型的对象，定义如下:

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "companiesAndLocations", propOrder = { "count", "normalizedLocations", "companyList", "companyMap", "modifiers",
        "modifiersMap", "companyCount", "navigators" })
public class CompaniesAndLocations extends BaseCompanies implements Viewable {

    @XmlElement(name = "normalizedLocations", required = false)
    protected List<NormalizedLocation> normalizedLocations;

    public List<NormalizedLocation> getNormalizedLocations() {
        if (normalizedLocations == null) {
            normalizedLocations = new ArrayList<NormalizedLocation>();
        }
        return normalizedLocations;
    }

}

BaseCompanies 定义了许多其他字段:

@XmlTransient
public abstract class BaseCompanies {

    @XmlElement(name = "modifiers", required = false)
    private List<Modifiers> modifiers;
....

我应该补充一点，我与应用程序中其他工作代码使用的方法略有不同。其他资源方法从用 @XmlRegistry 注释的 ObjectFactory 获取它们的对象。 .不过，我认为这没有必要。我已经看到其他代码直接实例化 JAXB 注释的 POJO，而不使用 ObjectFactory 工厂。

有任何想法吗？

Answer 1

JAX-RS 尽其所能引导 JAXBContext，但它并不总是能够访问它需要了解的所有类。为此，您可以实现 ContextResolver。这使您可以完全控制 JAXBContext 的创建方式。下面是它的外观示例:

package org.example.order;

import javax.ws.rs.Produces;
import javax.ws.rs.ext.ContextResolver;
import javax.ws.rs.ext.Provider;
import javax.xml.bind.JAXBContext;
import javax.xml.transform.Source;
import javax.xml.transform.stream.StreamSource;

import org.eclipse.persistence.jaxb.JAXBContextFactory;

@Provider
@Produces({"application/xml", "application/json"})
public class PurchaseOrderContextResolver implements ContextResolver<JAXBContext> {

    private JAXBContext jaxbContext;

    public PurchaseOrderContextResolver() {
        try {
            // Bootstrap your JAXBContext will all necessary classes
            jaxbContext = JAXBContext.newInstance(PurchaseOrder.class);
        } catch(Exception e) {
            throw new RuntimeException(e);
        }
    }

    public JAXBContext getContext(Class<?> clazz) {
        if(PurchaseOrder.class == clazz) {
            return jaxbContext;
        }
        return null;
    }

}