java如何正确使用Jackson的JaxbAnnotationIntrospector？

3 周，4 日 Questions & Answers 11894

我能够使用@JacksonXmlProperty注释和默认映射器配置生成我想要的XML。但我的类是由maven-jaxb2-plugin生成的，并且已经有@xmltattribute注释。当我尝试使用JaxbAnnotationIntrospector时，它将属性序列化为子元素。我做错了什么

预期输出：<problem xmlns="" id="aaa"><description>test</description></problem>（可通过testGenerateXmlCorrect重复）

实际输出：<problem xmlns=""><id>aaa</id><description>test</description></problem>（可与TestGenerateXmlError重复）

我还可以使用JAXB生成预期的XML，但问题是如何使用JaxbAnnotationIntrospector与Jackson一起生成

Junit测试：

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlProperty;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlRootElement;
import com.fasterxml.jackson.module.jaxb.JaxbAnnotationIntrospector;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlRootElement;
import org.junit.Test;

public class JaxbAttributeTest {
private static final org.slf4j.Logger log = org.slf4j.LoggerFactory.getLogger(JaxbAttributeTest.class);

@XmlRootElement(name="problem")
public static class ProblemJaxb {
    @XmlAttribute(name="id")     
    public String id;
    public String description;
}

@Test
public void testGenerateXmlWrong() throws JsonProcessingException {
    ProblemJaxb problem = new ProblemJaxb();
    problem.id = "aaa";
    problem.description = "test";
    XmlMapper xmlMapper = new XmlMapper();
    xmlMapper.setAnnotationIntrospector(new JaxbAnnotationIntrospector(xmlMapper.getTypeFactory()));
    log.debug("ProblemJaxb: {}", xmlMapper.writeValueAsString(problem)); 
}

@JacksonXmlRootElement(localName="problem")
public static class ProblemJackson {
    @JacksonXmlProperty(isAttribute=true)
    public String id;
    public String description;
}

@Test
public void testGenerateXmlCorrect() throws JsonProcessingException {
    ProblemJackson problem = new ProblemJackson();
    problem.id = "aaa";
    problem.description = "test";
    XmlMapper xmlMapper = new XmlMapper();
    log.debug("ProblemJackson: {}", xmlMapper.writeValueAsString(problem));
}

}

类路径包括：

com。fasterxml。杰克逊。core:jackson core:jar:2.3.2
com。fasterxml。杰克逊。核心：jackson注释：jar:2.3.2
com。fasterxml。杰克逊。核心：jackson数据绑定：jar:2.3.2
com。fasterxml。杰克逊。dataformat:jackson dataformat xml:jar:2.3.2
com。fasterxml。杰克逊。模块：jackson模块jaxb注释：jar:2.3.2
组织。科德豪斯。woodstox:stax2 api:jar:3.1.1
javax。xml。流：staxapi:jar:1.0-2
com。太阳xml。bind:jaxb impl:jar:2.2.7
com。太阳xml。bind:jaxb-core:jar:2.2.7
javax。xml。bind:jaxb-api:jar:2.2.7

顺便说一句，我还尝试使用以下配置XmlMapper：

xmlMapper.getSerializationConfig().with(new JaxbAnnotationIntrospector(xmlMapper.getTypeFactory()));

但这产生了更糟糕的输出，因为根元素名称不正确：<ProblemJaxb xmlns=""><id>aaa</id><description>test</description></ProblemJaxb>

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java如何正确使用Jackson的JaxbAnnotationIntrospector？

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