java - HTTP URL connection: display specific line of website source code -

im building first application , im kind of new but:

i'm sending http request using httpurlconnection. here's activity code:

    import java.io.bufferedreader; import java.io.ioexception; import java.io.inputstream; import java.io.inputstreamreader; import java.net.httpurlconnection; import java.net.url; import java.net.urlconnection; import android.app.activity; import android.os.asynctask; import android.os.bundle; import android.view.view; import android.view.view.onclicklistener; import android.widget.button; import android.widget.textview;  public class httpgetservletactivity3 extends activity implements     onclicklistener { button button; textview outputtext;  public static final string url =     "url";  /** called when activity first created. */ @override public void oncreate(bundle savedinstancestate) {     super.oncreate(savedinstancestate);     setcontentview(r.layout.main);      findviewsbyid();      button.setonclicklistener(this); }  private void findviewsbyid() {     button = (button) findviewbyid(r.id.button);     outputtext = (textview) findviewbyid(r.id.outputtxt); }  public void onclick(view view) {     getxmltask task = new getxmltask();     task.execute(new string[] { url }); }  private class getxmltask extends asynctask<string, void, string> {     @override     protected string doinbackground(string... urls) {         string output = null;         (string url : urls) {             output = getoutputfromurl(url);         }         return output;     }      private string getoutputfromurl(string url) {         stringbuffer output = new stringbuffer("");         try {             inputstream stream = gethttpconnection(url);             bufferedreader buffer = new bufferedreader(                     new inputstreamreader(stream));             string s = "";             while ((s = buffer.readline()) != null)                 output.append(s);         } catch (ioexception e1) {             e1.printstacktrace();         }         return output.tostring();     }      // makes httpurlconnection , returns inputstream     private inputstream gethttpconnection(string urlstring)             throws ioexception {         inputstream stream = null;         url url = new url(urlstring);         urlconnection connection = url.openconnection();          try {             httpurlconnection httpconnection = (httpurlconnection) connection;             httpconnection.setrequestmethod("get");             httpconnection.connect();              if (httpconnection.getresponsecode() == httpurlconnection.http_ok) {                 stream = httpconnection.getinputstream();             }         } catch (exception ex) {             ex.printstacktrace();         }         return stream;     }      @override     protected void onpostexecute(string output) {         outputtext.settext(output);     } } } 

this code me display website's source code, , it's working !! if want specific line of source code? example line 60? tried change following buuferedreader block:

bufferedreader buffer = new bufferedreader(                     new inputstreamreader(stream));             string s = "";             while ((s = buffer.readline()) != null)                 output.append(s); 

and wrote following linereader instead didnt work:

linenumberreader lnr = new linenumberreader(new inputstreamreader(stream));                string s = "";             if (lnr.getlinenumber() == 60){                 output.append(s);             } 

is way or missing ?

you can use setlinenumber method;

lnr.setlinenumber(60) 

and then, can read line!