The byte[] that is returned by FileUtils.readFileToByteArray is only the file contents, nothing else.

You should create your own class that is Serializable that includes two fields: a byte[] for the file contents, and a java.io.File that has everything else you need. You then serialize/deserialize your class into byte[], which is transmitted.

No. You can take a guess at a mimetype from the content data itself, but the filename is not in there.

The array of bytes doesn't contain the file name. You'd have to send it separately. You can call getFileName on your path, and then turn that into a byte array using getBytes() on the resulting string.

String fileName = path.getFileName();
byte[] fileNameBytes = fileName.getBytes();

You can then send this first and read it on the other end. Note, this wont contain the whole path, only the name of the file (music.mp3 in your case).

By the way, are you sure you want to be using UDP? What if you lose a packet or two when the data is being transferred? How do you detect that on the server?

This is late but if anyone is still looking for an answer.

ByteArrayResource doesn't have any filename by default. See the super class method AbstractResource.getFileName(), it returns null. So until you set the name explicitly while creating ByteArrayResource, it will remain null.

You can achieve this like

ByteArrayResource resource = new ByteArrayResource(bytearray) {
        @Override
        public String getFilename() {
            return "somename";
        };
    }

Now how to use the actual file name instead of "somename". If you're using Spring, you can define a method in your controller accepting Multipart as an argument.

public ResponseEntity<byte[]> handleFileOperation(Multipart file) {}

Once you get the MultipartFile instance, you can use it the way you want.

You can't get a filename from a binary data received from stream or socket, unless you know something about these bytes, for example

First ten bytes represent filename in ASCII encoding.

If you have no such guarantees, no chance.

From JDK 7 you can use Files.readAllBytes(Path).

Example:

import java.io.File;
import java.nio.file.Files;

File file;
// ...(file is initialised)...
byte[] fileContent = Files.readAllBytes(file.toPath());