How to instantiate java.io.File object without knowing file extension?

Question

I store file names of images in my database(user_id.extension, for example 14.png).

Suppose, I need to retrieve a file and write it to stream. But how can I create the java.io.File object without knowing file extension(images can be jpg, png and so on)? Now in my database there are only images with .png extension and I do this:

File file = new File(pathToAvatarFolder + File.separator + user.getId() + ".png");

Is there an elegant solution for this? Only brute force solution comes to my mind:

List<String> imageExtensions = new ArrayList<>();
imageExtensions.add(".png");
imageExtensions.add(".jpg");
imageExtensions.add(".jpeg");

File file;
for(int i = 0; i < imageExtensions.size(); i++) {
    file = new File(pathToAvatarFolder + File.separator + user.getId() + imageExtensions.get(i));
    if (file.exists()) {
        break;
    }

    file = null;
}

Just to be clear, I need to instance java.io.File object using pathname without extension.

At the long run store the file extension / mime type too. Your solution probably is faster than doing a directory scan with a file filter. — Joop Eggen
– Joop Eggen, Commented Jan 2, 2018 at 11:02
As Joop mentioned save the file extension in your database, too. Or alternatively change the avatar fileext to a fixed value. Anyway I would recommend you to generate a random filename on upload. This prevents a lot of attack scenarios. — JMax
– JMax, Commented Jan 2, 2018 at 12:55

Sai Kishore · Accepted Answer · 2018-01-02 11:07:39Z

1

import java.nio.*;
import java.nio.file.*;

Files.probeContentType((new File("filename.ext")).toPath());

This returns the type of the files. For png, it will be image/png.

answered Jan 2, 2018 at 11:07

Sai Kishore

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3 Comments

JMax Over a year ago

This won't work. The OP is missing the the concrete path to the file, therefore it's content can't be checked.

MikeJRamsey56 Over a year ago

"Just to be clear, I need to instance java.io.File object using pathname without extension." @salkisore The OP claims to have the path.

Sai Kishore Over a year ago

I used this to determine whether the given file is an image or not. if(Files.probeContentType((new File("filename.ext")).toPath()).startsWith("image")) {...} Using this condition, we can later instantiate the File.

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