java - 嵌套元素列表的 JAXB 注释

Question

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>

我有以下 java 类将其映射到：

@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

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.

我需要参数来同时具有'attr'属性和内容，那么我做错了什么？

Answer 1

默认情况下，JAXB (JSR-222)实现将公共属性（get/set 方法）和带注释的字段视为映射（和单独的）。默认映射是@XmlElement这样您的属性将被视为以这种方式映射。

解决方案#1

由于您正在注释需要@XmlAccessorType(XmlAccessType.FIELD)在类中添加的字段。

@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

您还需要使用属性@XmlElement上的注释mappings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;
    }
}