Create a file object from a resource path to an image in a jar file

Question

I need to create a File object out of a file path to an image that is contained in a jar file after creating a jar file. If tried using:

URL url = getClass().getResource("/resources/images/image.jpg");
File imageFile = new File(url.toURI());

but it doesn't work. Does anyone know of another way to do it?

score 4 · Answer 1 · answered Apr 09 '12 at 16:26

To create a file on Android from a resource or raw file I do this:

try{
  InputStream inputStream = getResources().openRawResource(R.raw.some_file);
  File tempFile = File.createTempFile("pre", "suf");
  copyFile(inputStream, new FileOutputStream(tempFile));

  // Now some_file is tempFile .. do what you like
} catch (IOException e) {
  throw new RuntimeException("Can't create temp file ", e);
}

private void copyFile(InputStream in, OutputStream out) throws IOException {
    byte[] buffer = new byte[1024];
    int read;
    while((read = in.read(buffer)) != -1){
      out.write(buffer, 0, read);
    }
}

Don't forget to close your streams etc

This is THE answer of all answers on this topic! – Jesse Jul 19 '12 at 14:47 — Jesse, Jul 19 '12 at 14:47

Will · Answer 2 · 2011-01-04T19:44:51.410

3

This should work.

String imgName = "/resources/images/image.jpg";
InputStream in = getClass().getResourceAsStream(imgName);
ImageIcon img = new ImageIcon(ImageIO.read(in));

edited Jan 04 '11 at 19:44

answered Jan 04 '11 at 19:11

Will

6,179
4
31
49

2

Quick fix: `getResource` returns a `URL`, `getResourceAsStream` returns an `InputStream` – Powerlord Jan 04 '11 at 19:15

score 2 · Accepted Answer · answered Jan 04 '11 at 19:59

2

Usually, you can't directly get a java.io.File object, since there is no physical file for an entry within a compressed archive. Either you live with a stream (which is best most in the cases, since every good API can work with streams) or you can create a temporary file:

    URL imageResource = getClass().getResource("image.gif");
    File imageFile = File.createTempFile(
            FilenameUtils.getBaseName(imageResource.getFile()),
            FilenameUtils.getExtension(imageResource.getFile()));
    IOUtils.copy(imageResource.openStream(),
            FileUtils.openOutputStream(imageFile));

answered Jan 04 '11 at 19:59

mhaller

14,122
1
42
61

1

What package is FilenameUtils in? – MBU Jan 04 '11 at 20:27
1

Note that code snippet would not work for either an applet or an app. deployed using JWS. Unless the resource requires editing (unlikely with an image), expanding it to a temporary file is usually not the way to go. – Andrew Thompson Jan 05 '11 at 01:38
https://stackoverflow.com/questions/4301329/java-class-getresource-returns-null – the_prole Jun 05 '20 at 23:09

score 1 · Answer 4 · answered Jan 04 '11 at 19:10

1

You cannot create a File object to a reference inside an archive. If you absolutely need a File object, you will need to extract the file to a temporary location first. On the other hand, most good API's will also take an input stream instead, which you can get for a file in an archive.

answered Jan 04 '11 at 19:10

Konstantin Komissarchik

28,879
6
61
61

1

"On the other hand, most good API's will also take an input stream instead, which you can get for a file in an archive." Specifically, you can use `getResourceAsStream` instead of `getResource` – Powerlord Jan 04 '11 at 19:14

Create a file object from a resource path to an image in a jar file

4 Answers4

Linked