Android 将inputStream传递给XmlParser时出现FileNotFound异常

我渴望android开发。我将SAX解析器用于xml解析器。我找不到这个例外的原因。 我尝试了getAsset（）方法。但它不起作用

xmlParser代码：：：

 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;
   }
 }

---

为什么要使用带有URL的FileInputStream？尝试：

 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;
   }
 }

换线

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

借

其中，callWebservice方法如下所示

 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;

    }

---

newfileinputstream（“http://64.85.165.53/dharatest/xmlarray.xml“”

？？？您需要先从该url下载.xml文件，然后才能读取/解析该xml。既然您有

url（“…”），为什么要使用此新方法呢？openStream（）

？@A.A在我的例子中，我使用的url与上述代码不兼容，也就是说，我使用的代码如图所示，几乎适用于每个url链接

 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;

    }

 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;
   }
   }