如何自定义Jersey JAXB序列化的XML输出

Question

我在这里有一些@ javax.xml.bind.annotation.Xml ...注释类用于RESt Web服务.Jersey是在一个Spring托管的Web容器中设置的,Web服务正在返回格式良好的xml.我们使用maven-enunciate-plugin来记录Web服务并为返回的xml文档创建xsd.我现在想将文档xsd文件用作返回的xml文件中的schemaLocation,以便xml验证不会抱怨缺少定义.如何为此配置XML序列化？

Answer 1

如果我没记错的话,我必须做一些事情才能将命名空间标识符正确地写入我生成的XML中.

1)创建了一个JaxbFactory,它配置并返回一个自定义编组器(和unmarshaller,BTW).我省略下面的getter /和解组设置......

//constructor
public JaxbFactory() throws Exception {
    context = JAXBContext.newInstance(ResourceDto.class);

    // Setup the marshaller
    marshaller = context.createMarshaller();
    marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);
    marshaller.setProperty(Marshaller.JAXB_SCHEMA_LOCATION, XmlMetadataConstants.XML_SCHEMA_LOCATION);  // this schema location is used in generating the schema-location property in the xml
}

2)泽西岛的"工厂级"不是"可见的".为了使它可见,我创建了一个MarshallerProvider.看起来像这样:

@Provider
public class ResourceJaxbMarshallerProvider implements ContextResolver<Marshaller> {
// injected by Spring
private ResourceJaxbFactory ResourceJaxbFactory;
private ResourceStatusJaxbFactory ResourceStatusJaxbFactory;


/*
 * ----------------------------------------
 * Setters (for Spring injected properties)
 * ----------------------------------------
 */
public void setResourceJaxbFactory(ResourceJaxbFactory ResourceJaxbFactory) {
    this.ResourceJaxbFactory = ResourceJaxbFactory;
}

public void setResourceStatusJaxbFactory(ResourceStatusJaxbFactory ResourceStatusJaxbFactory) {
    this.ResourceStatusJaxbFactory = ResourceStatusJaxbFactory;
}

/*
 * ------------------------
 * Interface Implementation
 * ------------------------
 */
public Marshaller getContext(Class<?> type) {
    if (type == ResourceDto.class)
        return ResourceJaxbFactory.getMarshaller();
    else if (type == ResourceStatusDto.class)
        return ResourceStatusJaxbFactory.getMarshaller();
    else
        return null;
}       
}

我使用Jersey/Spring Servlet将Jersey连接到Spring,因此任何由Spring创建的@Provider类都会被Jersey自动识别.在我的Spring applicationContext.xml中,我所要做的就是实例化资源提供者.反过来,它会从工厂中抓住编组员.

3)我发现关键的另一件事是我必须在包含我的资源的根包中创建一个package-info.java文件.看起来像这样:

/*
 * Note that this file is critical for ensuring that our ResourceDto object is
 * marshalled/unmarshalled with the correct namespace.  Without this, marshalled
 * classes produce XML files without a namespace identifier
 */
@XmlSchema(namespace = XmlMetadataConstants.XML_SCHEMA_NAMESPACE, elementFormDefault = XmlNsForm.QUALIFIED) 
package com.yourcompany.resource;

import javax.xml.bind.annotation.XmlNsForm;

至少我认为这是我需要做的一切,我不记得每件作品.我确实记得,package-info.java是最后一个让它全部融合在一起的关键因素.

希望有所帮助.我花了太多时间来挖掘所有这些信息.在我希望它进行适当的xml架构验证(以及架构无效输入的正确错误报告)之前,Jersey很诱人.一旦我开始沿着这条路开始,泽西从脑死亡变得容易到相当困难.大多数困难是从网上各种帖子中删除所有细节.希望这将有助于您更快,更快.:-)