CODEWITHSUNDEEP
javastream
Stripies
Stripies

Reputation: 1267

Reading several byte[] from DataInputStream

What I need to do is send multiple files using DataStreams. I'm doing this by sending the name of the file, and then the file's bytes. I need to send an undetermined number of files though. Here is the DataOutputStream code.

            out.writeUTF(path);
            out.write(Files.readAllBytes(file.toPath()));

It does that for each file that needs to be sent. But I don't know how to read it correctly with DataInputStream. This is what I have so far.

    while (in.available() != 0) {
        String path = in.readUTF();
        byte bytes = in.readByte();
    }

Obviously it wouldn't work, since it is only reading one byte. But I don't know how to make it read all of the bytes. Since there are several files being sent, available() would only equal 0 when the end of all the files are read, I think. Any help is greatly appreciated.

Something I completely forgot to mention, I want to be able to send a large file without running out of memory, and I don't think this would work. I think I would need to use a buffer, but I don't know what class supports that with files.

Upvotes: 1

Views: 2979

Answers (2)

Hiro2k
Hiro2k

Reputation: 5587

Anytime you send variable length messages you need some way to mark the beginning and end of each method.

 List<File> files = someListOfFilesYouWantToSend;
 out.writeInt(files.size());
 for(File file : files){
   out.writeUTF(path);
   out.writeLong(file.getTotalSpace());
   out.write(Files.readAllBytes(file.toPath()));
 }

Then to read it you would do something like this

int filesToRead = in.readInt();    
for(int i = 0; i < filesToRead; i++){
  String path = in.readUTF();
  long bytesToRead = in.readLong();      
  FileOutputSteam fos = new FileOutputStream(path);

  byte[] buffer = new byte[1024];
  while(bytesToRead > 0){
    bytesRead = in.read(buffer,0,bytesToRead > buffer.length ? buffer.length : bytesToRead);
    bytesToRead -= bytesRead;
    fos.write(buffer);
  }
}

Upvotes: 2

smichak
smichak

Reputation: 4958

That's not the way to do it... Why won't you simply archive all the files you want to send in an archive (like a JAR or ZIP)? On the receiving side you can extract the archive. Java has a built-in JAR implementation (in package java.util.jar) that you can use.

Upvotes: 0

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