用于嵌套元素列表的JAXB注释

I have the following XML:

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>

I have the following java classes to map it to:

我有以下java类来映射到:

@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;
    }
}

When I try to unmarshall it with

当我试图解开它的时候

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

I get this error

我得到这个错误

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

I need parameter to have both the 'attr' attribute and content, so what am I doing wrong?

我需要参数来拥有“attr”属性和内容，那么我做错了什么呢?

1 个解决方案

#1

By default JAXB (JSR-222) implementations consider public properties (get/set methods) and annotated fields as mapped (and separate). The default mapping is @XmlElement so your properties will be considered as mapped this way.

默认情况下，JAXB (JSR-222)实现考虑公共属性(get/set方法)和注释字段作为映射(和独立)。默认映射是@XmlElement，因此您的属性将被认为是这样映射的。

Solution #1

解决方案1

Since you are annotating the fields you need to add @XmlAccessorType(XmlAccessType.FIELD) on your classes.

由于您正在注释需要在类上添加@XmlAccessorType(XmlAccessType.FIELD)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;
    }
}

Solution #2

解决方案2

Annotate the get (or set) methods.

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;
    }
}

For More Information

的更多信息

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

UPDATE

更新

You will also need to use the @XmlElenent annotation on the mappings property to specify the element name should be mapping.

您还需要在mappings属性上使用@XmlElenent注释来指定应该映射的元素名。

@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;
    }
}

#1