如何将XML文件转换为json

Question

我想将我的xml文件转换为json ...，但是下面的代码显示了空指针异常..我不知道出了什么问题。

public class Xmljson {  
    private URL url = null;
    private InputStream inputStream = null;  

    public void getXMLfromJson() {
        try{
            url = Xmljson.class.getClassLoader().getResource("datafile.xml");
            inputStream = url.openStream();
            String xml = IOUtils.toString(inputStream);
            JSON objJson = new XMLSerializer().read(xml);
            System.out.println("JSON data : " + objJson);
        }catch(Exception e){
            e.printStackTrace();
        }finally{
           try {
                if (inputStream != null) {
                    inputStream.close();
                }
                url = null;
            } catch (IOException ex) {}
        }
    }   

    public static void main(String[] args) {
        new Xmljson().getXMLfromJson();
    }
}

在这里显示异常

         url = Xmlto.class.getClassLoader().getResource("data");
            inputStream = url.openStream();

NPE IS

java.lang.NullPointerException
at pkg.news.Xmlto.getXMLfromJson(Xmlto.java:19)
at pkg.news.Xmlto.main(Xmlto.java:35)

引自

http://tutorial4java.blogspot.in/2013/04/xml-to-json-conversion.html

Answer 1

检查一下...这很完美...

  public class Xml2json {

    static String line="",str="";
    public static void main(String[] args) throws JSONException, IOException {
        String link = "data.xml";
        BufferedReader br = new BufferedReader(new FileReader(link));
        while ((line = br.readLine()) != null) 
        {   
            str+=line;  
        }
        JSONObject jsondata = XML.toJSONObject(str);
        System.out.println(jsondata);
    }
}

Answer 2

它应该工作

url = XMLjson.class.getClassLoader().getResource("datafile.xml");

您是否也添加了这些语句，

import java.io.InputStream;  
import java.net.URL;  
import net.sf.json.JSON;  
import net.sf.json.xml.XMLSerializer;  
import org.apache.commons.io.IOUtils;  

也在这里看看http://tutorial4java.blogspot.in/2013/04/xml-to-json-conversion.html