Question

我有以下XML

<search ver="3.0">
    <loc id="ARBA0009" type="1">Buenos Aires, Argentina</loc>
    <loc id="BRXX1283" type="1">Buenos Aires, Brazil</loc>
    <loc id="ARDF0127" type="1">Aeroparque Buenos Aires, Argentina</loc>
    <loc id="MXJO0669" type="1">Concepcion De Buenos Aires, Mexico</loc>
    <loc id="MXPA1785" type="1">San Nicolas De Buenos Aires, Mexico</loc>
    <loc id="ARBA0005" type="1">Balcarce, Argentina</loc>
    <loc id="ARBA0008" type="1">Bragado, Argentina</loc>
    <loc id="ARBA0010" type="1">Campana, Argentina</loc>
    <loc id="ARBA0016" type="1">Chascomus, Argentina</loc>
    <loc id="ARBA0019" type="1">Chivilcoy, Argentina</loc>
</search>

和一个城市班级

public class City {

    private String  id;
    private Integer type;
    private String  name;

    // getters & setters...
}

我尝试了以下别名来解析XML

xStream.alias("search", List.class);
xStream.alias("loc", City.class);
xStream.useAttributeFor("id", String.class);
xStream.useAttributeFor("type", Integer.class);

但我无法弄清楚如何设置“loc”标签的值，如果我尝试使用XML转换City对象

<search>
    <loc id="ARBA0001" type="1">
        <name>Buenos Aires</name>
    </loc>
</search>

当我真的需要这个时

<search>
    <loc id="ARBA0001" type="1">Buenos Aires</loc>
</search>

然后，如果我尝试将XML解析为City对象，我会得到带有空值的字段“name”。

有谁知道如何设置te正确的别名来做到这一点？提前谢谢。

Answer 1

我终于找到了解决方案，转换器解决了这个问题，这里是代码

public class CityConverter implements Converter {

    public void marshal(Object value, HierarchicalStreamWriter writer, 
                                                               MarshallingContext context) {
        City city = (City) value;
        writer.addAttribute("id", city.getId());
        writer.addAttribute("type", city.getType().toString());
        writer.setValue(city.getName());
    }

    public Object unmarshal(HierarchicalStreamReader reader, UnmarshallingContext context) {
        City city = new City();
        city.setName(reader.getValue());
        city.setId(reader.getAttribute("id"));
        city.setType(reader.getAttribute("type"));
        return city;
    }

    public boolean canConvert(Class clazz) {
        return clazz.equals(City.class);
    }

}

在设置别名的部分，我还设置了CityConverter

xStream.registerConverter(new CityConverter());
xStream.alias("search", List.class);
xStream.alias("loc", City.class);

一切正常：）

Answer 2

我发帖说希望它可以帮助别人，因为我花了很长时间才找到它... http://fahdshariff.blogspot.com/2011/12/using-xstream-to-map-single-element.html

答案是使用@XStreamConverter - ToAttributedValueConverter

@XStreamAlias("error")
@XStreamConverter(value=ToAttributedValueConverter.class, strings={"message"})
public class Error {

  String message;

  ...

有许多有趣的转换器提供各种有用的功能...... http://x-stream.github.io/converters.html

Answer 3

XStream看起来有点复杂，您可以在JAXB中执行以下操作：

public class City { 

    @XmlAttribute private String  id; 
    @XmlAttribute private Integer type; 
    @XmlValue private String  name; 

    // getters & setters... 
}

Answer 4

@XStreamConverter(value= ToAttributedValueConverter.class, strings={"name"})
public class City { 

    @XStreamAsAttribute @XStreamAlias("id") private String  id; 
    @XStreamAsAttribute @XStreamAlias("type") private Integer type; 
    private String  name; 

    // getters & setters... 
}

XStream同时解析属性和值

4 个答案: