How to download and save a file from Internet usin

Question

There is an online file (such as http://www.example.com/information.asp) I need to grab and save to a directory. I know there are several methods for grabbing and reading online files (URLs) line-by-line, but is there a way to just download and save the file using Java?

Answer 1

import java.io.*;
import java.net.*;

public class filedown {
    public static void download(String address, String localFileName) {
        OutputStream out = null;
        URLConnection conn = null;
        InputStream in = null;

        try {
            URL url = new URL(address);
            out = new BufferedOutputStream(new FileOutputStream(localFileName));
            conn = url.openConnection();
            in = conn.getInputStream();
            byte[] buffer = new byte[1024];

            int numRead;
            long numWritten = 0;

            while ((numRead = in.read(buffer)) != -1) {
                out.write(buffer, 0, numRead);
                numWritten += numRead;
            }

            System.out.println(localFileName + "\t" + numWritten);
        } 
        catch (Exception exception) { 
            exception.printStackTrace();
        } 
        finally {
            try {
                if (in != null) {
                    in.close();
                }
                if (out != null) {
                    out.close();
                }
            } 
            catch (IOException ioe) {
            }
        }
    }

    public static void download(String address) {
        int lastSlashIndex = address.lastIndexOf('/');
        if (lastSlashIndex >= 0 &&
        lastSlashIndex < address.length() - 1) {
            download(address, (new URL(address)).getFile());
        } 
        else {
            System.err.println("Could not figure out local file name for "+address);
        }
    }

    public static void main(String[] args) {
        for (int i = 0; i < args.length; i++) {
            download(args[i]);
        }
    }
}

Answer 2

public void saveUrl(final String filename, final String urlString)
        throws MalformedURLException, IOException {
    BufferedInputStream in = null;
    FileOutputStream fout = null;
    try {
        in = new BufferedInputStream(new URL(urlString).openStream());
        fout = new FileOutputStream(filename);

        final byte data[] = new byte[1024];
        int count;
        while ((count = in.read(data, 0, 1024)) != -1) {
            fout.write(data, 0, count);
        }
    } finally {
        if (in != null) {
            in.close();
        }
        if (fout != null) {
            fout.close();
        }
    }
}

You'll need to handle exceptions, probably external to this method.

Answer 3

Personally, I've found Apache's HttpClient to be more than capable of everything I've needed to do with regards to this. Here is a great tutorial on using HttpClient

Answer 4

This is another java7 variant based on Brian Risk's answer with usage of try-with statement:

public static void downloadFileFromURL(String urlString, File destination) throws Throwable {

      URL website = new URL(urlString);
      try(
              ReadableByteChannel rbc = Channels.newChannel(website.openStream());
              FileOutputStream fos = new FileOutputStream(destination);  
              ){
          fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);
      }

  }

Answer 5

Give Java NIO a try:

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);

Using transferFrom() is potentially much more efficient than a simple loop that reads from the source channel and writes to this channel. Many operating systems can transfer bytes directly from the source channel into the filesystem cache without actually copying them.

Check more about it here.

Note: The third parameter in transferFrom is the maximum number of bytes to transfer. Integer.MAX_VALUE will transfer at most 2^31 bytes, Long.MAX_VALUE will allow at most 2^63 bytes (larger than any file in existence).

Answer 6

You can do this in 1 line using netloader for Java:

new NetFile(new File("my/zips/1.zip"), "https://example.com/example.zip", -1).load(); //returns true if succeed, otherwise false.