SimpleXML Как перенести это в код Android?

Question

Я хочу использовать retrofit2 с simplexml, чтобы преобразовать http-запрос. Это android код. У меня проблемы с преобразованием xml. Как это сделать? Как я могу написать класс?

<DataSet xmlns="http://WebXml.com.cn/"> 
    <xs:schema xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" id="Dictionary">...</xs:schema> 
    <diffgr:diffgram xmlns:msdata="urn:schemas-microsoft-com:xml-msdata" xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1"> 
     <Dictionary xmlns=""> 
      <Trans diffgr:id="Trans1" msdata:rowOrder="0"> 
       <WordKey>hello</WordKey> 
       <Pron>'heləu, he'ləu</Pron> 
       <Info/> 
       <Translation>int.（见面打招呼或打电话用语）喂,哈罗</Translation> 
       <Mp3>1059.mp3</Mp3> 
      </Trans> 
      <Sentence diffgr:id="Sentence1" msdata:rowOrder="0"> 
       <Orig> 
        She actually condescended to say hello to me in the street today. 
       </Orig> 
       <Trans>她今天在街上竟能屈尊跟我打招呼.</Trans> 
      </Sentence> 
      <Sentence diffgr:id="Sentence2" msdata:rowOrder="1"> 
       <Orig> 
        I said hello to her, but she ignored me completely! 
       </Orig> 
       <Trans>我向她打招呼, 可她根本不理我!</Trans> 
      </Sentence> 
      <Sentence diffgr:id="Sentence3" msdata:rowOrder="2"> 
       <Orig>Hello there, what a coincidence!</Orig> 
       <Trans>你好，真巧啊!</Trans> 
      </Sentence> 
     </Dictionary> 
    </diffgr:diffgram> 
</DataSet> 

Answer 1

Найдено решение, пропуская xs:schema и diffgr:diffgram так:

@Root(name = "DataSet", strict = false) 
@Namespace(reference = "http://WebXml.com.cn/") 
public class DataSet { 

    @Path("diffgr:diffgram[1]") 
    @Element(name = "Dictionary", required = false) 
    private Dictionary dictionary; 

}