FileNotFound Exception while passing inputStream to XmlParser

Question

I am begginer to android development. I am using SAX parser for xml parser. I couldn't find the reason for this exeption. I tried getAsset() method. but it didn't worked.

xmlParser code :::

 public class XMLParser {

     public static Country parseCountry(InputStream is) {
        try {
         Country country= new Country(null, null, null);

         XMLReader xmlReader =  SAXParserFactory.newInstance().newSAXParser().getXMLReader();
         XMLHandler xmlHandler = new XMLHandler();
         xmlReader.setContentHandler(xmlHandler);     
         xmlReader.parse(new InputSource(new FileInputStream(http://64.85.165.53/dharatest/xmlarray.xml));
         country = xmlHandler.getParsedCountryData();

        } catch(ParserConfigurationException pce) { 
               Log.e("SAX XML", "sax parse error", pce); 
        } catch(SAXException se) { 
               Log.e("SAX XML", "sax error", se);       
        } catch(IOException ioe) { 
               Log.e("SAX XML", "sax parse io error", ioe); 
        }     
       return country;
   }
 }

`new FileInputStream("http://64.85.165.53/dharatest/xmlarray.xml")` ??? — Raghav, Jun 27 '12 at 12:59
you need to download your .xml file from that url first and then u can read/parse that XML. — Aqif Hamid, Jun 27 '12 at 13:02

score 2 · Accepted Answer · answered Jun 27 '12 at 13:03

Why are you using a FileInputStream with a URL? Try:

 public class XMLParser {

     public static Country parseCountry(InputStream is) {
        try {
         Country country= new Country(null, null, null);

         XMLReader xmlReader =  SAXParserFactory.newInstance().newSAXParser().getXMLReader();
         XMLHandler xmlHandler = new XMLHandler();
         xmlReader.setContentHandler(xmlHandler);     
         xmlReader.parse(new InputSource(new URL("http://64.85.165.53/dharatest/xmlarray.xml").openStream());
         country = xmlHandler.getParsedCountryData();

        } catch(ParserConfigurationException pce) { 
               Log.e("SAX XML", "sax parse error", pce); 
        } catch(SAXException se) { 
               Log.e("SAX XML", "sax error", se);       
        } catch(IOException ioe) { 
               Log.e("SAX XML", "sax parse io error", ioe); 
        }     
       return country;
   }
 }

Jitendra Patidar · Answer 2 · 2012-06-28T09:37:13.310

 public class XMLParser {

     public static Country parseCountry(InputStream is) {
         try {
         Country country= new Country(null, null, null);

       XMLReader xmlReader =SAXParserFactory.newInstance().newSAXParser().getXMLReader();
     XMLHandler xmlHandler = new XMLHandler();
     xmlReader.setContentHandler(xmlHandler);     
     xmlReader.parse(new InputSource(getAssets().open("data.xml"));
     country = xmlHandler.getParsedCountryData();

    } catch(ParserConfigurationException pce) { 
           Log.e("SAX XML", "sax parse error", pce); 
    } catch(SAXException se) { 
           Log.e("SAX XML", "sax error", se);       
    } catch(IOException ioe) { 
           Log.e("SAX XML", "sax parse io error", ioe); 
    }     
  return country;
   }
   }

score -1 · Answer 3 · answered Jun 27 '12 at 13:10

change your line

xmlReader.parse(new InputSource(new FileInputStream(http://64.85.165.53/dharatest/xmlarray.xml));

by

 xmlReader.parse(new InputSource(callWebservice("http://64.85.165.53/dharatest/xmlarray.xml")));

where callWebservice method is as shown below

 private InputStream callWebservice(String serviceURL) {
        HttpClient client=new DefaultHttpClient();
        HttpGet getRequest=new HttpGet();

           try {
             // construct a URI object
             getRequest.setURI(new URI(serviceURL));
              } catch (URISyntaxException e) {
                Log.e("URISyntaxException", e.toString());
               }

            // buffer reader to read the response
           BufferedReader in=null;
           // the service response
          HttpResponse response=null;
             try {
                  // execute the request
                  response = client.execute(getRequest);
               } catch (ClientProtocolException e) {
                 Log.e("ClientProtocolException", e.toString());
               } catch (IOException e) {
                 Log.e("IO exception", e.toString());
            }
      if(response!=null)
            try {
                return response.getEntity().getContent();
            } catch (IllegalStateException e) {
                // TODO Auto-generated catch block

                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block

                e.printStackTrace();
            }
        else
          return null;
        return null;

    }

why use this new method while you have `URL("...").openStream()`? — ColdFire, Jun 27 '12 at 13:14
@A.A in my case the url which i have used is not works with the above code and that's y i have used code as shown aboe which works almost for each and every url links — Khan, Jun 27 '12 at 13:17

FileNotFound Exception while passing inputStream to XmlParser

3 Answers3