我有以下 XML:

<mappings>
    <mapping>
        <parameter attr = "value">asdas</parameter>
        <parameter attr = "value2">d123asdsad</parameter>
        <parameter attr = "value3">0</parameter>
    </mapping>
    <mapping>
        <parameter attr = "value">23123s</parameter>
        <parameter attr = "value2">qwerty</parameter>
        <!-- more parameter elements -->
    </mapping>
    <!-- more mapping elements -->
</mappings>

我将其映射到以下 java 类:

@XmlRootElement(name = "mappings")
public class Mappings { 
    @XmlElement(name = "mapping")
    private List<Mapping> mMappings;

    public List<Mapping> getMappings() {
        return mMappings;
    }

    public void setMappings(List<Mapping> aMappings) {
        this.mMappings = aMappings;
    }
}

public class Mapping {
    @XmlElement(name = "parameter")
    private List<Parameter> mParameters;

    public List<Parameter> getParameters() {
        return mParameters;
    }

    public void setParameters(List<Parameter> aParameters) {
        this.mParameters = aParameters;
    }
}

public class Parameter {
    @XmlAttribute(name = "attr")
    private String mName;

    @XmlValue
    private String mValue;

    public String getName() {
        return mName;
    }

    public void setName(String aName) {
        this.mName = aName;
    }

    public String getValue() {
        return mValue;
    }

    public void setValue(String aValue) {
        this.mValue = aValue;
    }
}

当我尝试用它解码时

JAXBContext context = JAXBContext.newInstance(BundleMappings.class);
Unmarshaller um = context.createUnmarshaller();
mappings = (BundleMappings)um.unmarshal(new File(myFile));

我收到这个错误

If a class has @XmlElement property, it cannot have @XmlValue property.

我需要参数同时具有“attr”属性和内容，那么我做错了什么？

最佳答案

默认 JAXB (JSR-222)实现将公共(public)属性(获取/设置方法)和带注释的字段视为映射(和分离)。默认映射为 @XmlElement，因此您的属性将被视为以这种方式映射。

解决方案#1

由于您正在注释字段，因此您需要在类中添加 @XmlAccessorType(XmlAccessType.FIELD)。

@XmlAccessorType(XmlAccessType.FIELD)
public class Parameter {
    @XmlAttribute(name = "attr")
    private String mName;

    @XmlValue
    private String mValue;

    public String getName() {
        return mName;
    }

    public void setName(String aName) {
        this.mName = aName;
    }

    public String getValue() {
        return mValue;
    }

    public void setValue(String aValue) {
        this.mValue = aValue;
    }
}

解决方案#2

注释 get(或 set)方法。

public class Parameter {
    private String mName;

     private String mValue;

    @XmlAttribute(name = "attr")
    public String getName() {
        return mName;
    }

    public void setName(String aName) {
        this.mName = aName;
    }

    @XmlValue
    public String getValue() {
        return mValue;
    }

    public void setValue(String aValue) {
        this.mValue = aValue;
    }
}

了解更多信息

• http://blog.bdoughan.com/2011/06/using-jaxbs-xmlaccessortype-to.html
• http://blog.bdoughan.com/2011/06/jaxb-and-complex-types-with-simple.html

更新

您还需要在 mappings 属性上使用 @XmlElement 注释来指定元素名称应该是 mapping。

@XmlRootElement(name = "mappings")
public class Mappings { 
    private List<Mapping> mMappings;

    @XmlElement(name="mapping")
    public List<Mapping> getMappings() {
        return mMappings;
    }

    public void setMappings(List<Mapping> aMappings) {
        this.mMappings = aMappings;
    }
}

关于java - 嵌套元素列表的 JAXB 注释，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/14202529/