Open URL In java

Question

How to open URL in java?

E.g i have https://www.iformbuilder.com/exzact/dataExcelFeed.php?
PAGE_ID=3175952&TABLE_NAME=_data399173_eff_sor&ANALYTIC=1&SINCE_DATE=2013-10-1

when we go to this URL it will download the file.But how to implement this thing in code?

I try this to open url so that file will be downloaded.But its not working.

URL url = new URL("https://www.iformbuilder.com/exzact/dataExcelFeed.php?PAGE_ID=3175952&TABLE_NAME=_data399173_eff_sor&ANALYTIC=1&SINCE_DATE=2013-10-16");
    HttpURLConnection urlCon = (HttpURLConnection) url.openConnection();
    System.out.println(urlCon);
    urlCon.connect();

I know something is wrong But i dont knw whats dat

I would suggest adding to the question things you've tried already. Did it not work the way you tried it? If you just want to know how to open and read a URL using Java, a Google search would help. Here's the first link that's returned when you query "URL java read" http://docs.oracle.com/javase/tutorial/networking/urls/readingURL.html — Shobit, Oct 16 '13 at 05:46
Possible duplicate of [How to download and save a file from Internet using Java?](https://stackoverflow.com/questions/921262/how-to-download-and-save-a-file-from-internet-using-java) — Robin Green, Nov 17 '18 at 06:34

score 2 · Answer 1 · edited May 23 '17 at 11:50

2

I have found this:

URL website = new URL("http://www.website.com/information.asp");
ReadableByteChannel rbc = Channels.newChannel(website.openStream());
FileOutputStream fos = new FileOutputStream("information.html");
fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);

In this article: How to download and save a file from Internet using Java?

I hope this helps you

edited May 23 '17 at 11:50

Community

1
1

answered Oct 16 '13 at 05:43

angel_navarro

1,757
10
11

Hanlet Escaño · Answer 2 · 2013-10-16T05:52:44.260

Try this:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;

class HttpHelperx
{
   public static String GET(String url)
   {
      String result = "";

      try 
      {
         URL navUrl = new URL(url);
         URLConnection con = (URLConnection)navUrl.openConnection();

         result = getContent(con);

      } 
      catch (MalformedURLException e) 
      {
         e.printStackTrace();
      } 
      catch (IOException e) 
      {
         e.printStackTrace();
      }

      return result;
   }

   public static String getContent(URLConnection con)
   {
      String result = "";
      if(con!=null)
      {
         BufferedReader br;
         try 
         {
            br = new BufferedReader(new InputStreamReader(con.getInputStream()));
            StringBuilder buffer = new StringBuilder();
            String aux = "";

            while ((aux = br.readLine()) != null)
            {
               buffer.append(aux);
            }
            result = buffer.toString();
            br.close();
         } 
         catch (IOException e) 
         {
            e.printStackTrace();
         }
      }

      return result;
   }
}

To use it:

public class HTTPHelperDriver
{
   public static void main(String[] args)
      throws Exception
   {
      String response = HttpHelperx.GET("http://www.google.com");
      System.out.println(response); 
   }
}

score 0 · Answer 3 · edited Jan 27 '22 at 21:25

0

You can use this code:

try {
    //specify the protocol along with the URL
    URI oURL = new URI("https://www.google.com/");
    desktop.browse(oURL);
} catch (URISyntaxException e1) {
    // TODO Auto-generated catch block
    e1.printStackTrace();
} catch (IOException e1) {
    // TODO Auto-generated catch block
    e1.printStackTrace();
}

edited Jan 27 '22 at 21:25

Dada

6,313
7
24
43

answered Jan 27 '22 at 16:29

Mpro

56
3

Open URL In java

3 Answers3