Java通过套接字发送和接收文件（字节[]）

Question

I am trying to develop a very simple client / server where the client converts a file to bytes, sends it to the server, and then converts the bytes back in to a file.

Currently the program just creates an empty file. I'm not a fantastic Java developer so any help much appreciated.

This is the server part that receives what the client sends.

ServerSocket serverSocket = null;

    serverSocket = new ServerSocket(4444);


    Socket socket = null;
    socket = serverSocket.accept();

    DataOutputStream out = new DataOutputStream(new BufferedOutputStream(socket.getOutputStream()));
    DataInputStream in = new DataInputStream(new BufferedInputStream(socket.getInputStream()));
    byte[] bytes = new byte[1024];

    in.read(bytes);
    System.out.println(bytes);

    FileOutputStream fos = new FileOutputStream("C:\\test2.xml");
    fos.write(bytes);

And here is the client part

Socket socket = null;
    DataOutputStream out = null;
    DataInputStream in = null;
    String host = "127.0.0.1";     

    socket = new Socket(host, 4444);
    out = new DataOutputStream(new BufferedOutputStream(socket.getOutputStream()));
    in = new DataInputStream(new BufferedInputStream(socket.getInputStream()));

    File file = new File("C:\\test.xml");
    //InputStream is = new FileInputStream(file);
    // Get the size of the file
    long length = file.length();
    if (length > Integer.MAX_VALUE) {
        System.out.println("File is too large.");
    }
    byte[] bytes = new byte[(int) length];

    //out.write(bytes);
    System.out.println(bytes);

    out.close();
    in.close();
    socket.close();

Answer 1

The correct way to copy a stream in Java is as follows:

int count;
byte[] buffer = new byte[8192]; // or 4096, or more
while ((count = in.read(buffer)) > 0)
{
  out.write(buffer, 0, count);
}

Wish I had a dollar for every time I've posted that in a forum.

Answer 2

Thanks for the help. I've managed to get it working now so thought I would post so that the others can use to help them.

Server:

public class Server {
    public static void main(String[] args) throws IOException {
        ServerSocket serverSocket = null;

        try {
            serverSocket = new ServerSocket(4444);
        } catch (IOException ex) {
            System.out.println("Can't setup server on this port number. ");
        }

        Socket socket = null;
        InputStream in = null;
        OutputStream out = null;
        
        try {
            socket = serverSocket.accept();
        } catch (IOException ex) {
            System.out.println("Can't accept client connection. ");
        }
        
        try {
            in = socket.getInputStream();
        } catch (IOException ex) {
            System.out.println("Can't get socket input stream. ");
        }

        try {
            out = new FileOutputStream("M:\\test2.xml");
        } catch (FileNotFoundException ex) {
            System.out.println("File not found. ");
        }

        byte[] bytes = new byte[16*1024];

        int count;
        while ((count = in.read(bytes)) > 0) {
            out.write(bytes, 0, count);
        }

        out.close();
        in.close();
        socket.close();
        serverSocket.close();
    }
}

and the Client:

public class Client {
    public static void main(String[] args) throws IOException {
        Socket socket = null;
        String host = "127.0.0.1";

        socket = new Socket(host, 4444);
        
        File file = new File("M:\\test.xml");
        // Get the size of the file
        long length = file.length();
        byte[] bytes = new byte[16 * 1024];
        InputStream in = new FileInputStream(file);
        OutputStream out = socket.getOutputStream();
        
        int count;
        while ((count = in.read(bytes)) > 0) {
            out.write(bytes, 0, count);
        }

        out.close();
        in.close();
        socket.close();
    }
}

Answer 3

Here is the server Open a stream to the file and send it overnetwork

import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.net.ServerSocket;
import java.net.Socket;

public class SimpleFileServer {

  public final static int SOCKET_PORT = 5501;
  public final static String FILE_TO_SEND = "file.txt";

  public static void main (String [] args ) throws IOException {
    FileInputStream fis = null;
    BufferedInputStream bis = null;
    OutputStream os = null;
    ServerSocket servsock = null;
    Socket sock = null;
    try {
      servsock = new ServerSocket(SOCKET_PORT);
      while (true) {
        System.out.println("Waiting...");
        try {
          sock = servsock.accept();
          System.out.println("Accepted connection : " + sock);
          // send file
          File myFile = new File (FILE_TO_SEND);
          byte [] mybytearray  = new byte [(int)myFile.length()];
          fis = new FileInputStream(myFile);
          bis = new BufferedInputStream(fis);
          bis.read(mybytearray,0,mybytearray.length);
          os = sock.getOutputStream();
          System.out.println("Sending " + FILE_TO_SEND + "(" + mybytearray.length + " bytes)");
          os.write(mybytearray,0,mybytearray.length);
          os.flush();
          System.out.println("Done.");
        } catch (IOException ex) {
          System.out.println(ex.getMessage()+": An Inbound Connection Was Not Resolved");
        }
        }finally {
          if (bis != null) bis.close();
          if (os != null) os.close();
          if (sock!=null) sock.close();
        }
      }
    }
    finally {
      if (servsock != null)
        servsock.close();
    }
  }
}

Here is the client Recive the file being sent overnetwork

import java.io.BufferedOutputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.Socket;

public class SimpleFileClient {

  public final static int SOCKET_PORT = 5501;
  public final static String SERVER = "127.0.0.1";
  public final static String
       FILE_TO_RECEIVED = "file-rec.txt";

  public final static int FILE_SIZE = Integer.MAX_VALUE;

  public static void main (String [] args ) throws IOException {
    int bytesRead;
    int current = 0;
    FileOutputStream fos = null;
    BufferedOutputStream bos = null;
    Socket sock = null;
    try {
      sock = new Socket(SERVER, SOCKET_PORT);
      System.out.println("Connecting...");

      // receive file
      byte [] mybytearray  = new byte [FILE_SIZE];
      InputStream is = sock.getInputStream();
      fos = new FileOutputStream(FILE_TO_RECEIVED);
      bos = new BufferedOutputStream(fos);
      bytesRead = is.read(mybytearray,0,mybytearray.length);
      current = bytesRead;

      do {
         bytesRead =
            is.read(mybytearray, current, (mybytearray.length-current));
         if(bytesRead >= 0) current += bytesRead;
      } while(bytesRead > -1);

      bos.write(mybytearray, 0 , current);
      bos.flush();
      System.out.println("File " + FILE_TO_RECEIVED
          + " downloaded (" + current + " bytes read)");
    }
    finally {
      if (fos != null) fos.close();
      if (bos != null) bos.close();
      if (sock != null) sock.close();
    }
  }    
}

Answer 4

To avoid the limitation of the file size , which can cause the Exception java.lang.OutOfMemoryError to be thrown when creating an array of the file size byte[] bytes = new byte[(int) length]; , instead we could do

    byte[] bytearray = new byte[1024*16];
    FileInputStream fis = null;
    try {

        fis = new FileInputStream(file);
        OutputStream output= socket.getOututStream();
        BufferedInputStream bis = new BufferedInputStream(fis);

        int readLength = -1;
        while ((readLength = bis.read(bytearray)) > 0) {
            output.write(bytearray, 0, readLength);

        }
        bis.close();
        output.close();
    }
    catch(Exception ex ){

        ex.printStackTrace();
    } //Excuse the poor exception handling...

Answer 5

Rookie, if you want to write a file to server by socket, how about using fileoutputstream instead of dataoutputstream? dataoutputstream is more fit for protocol-level read-write. it is not very reasonable for your code in bytes reading and writing. loop to read and write is necessary in java io. and also, you use a buffer way. flush is necessary. here is a code sample: http://www.rgagnon.com/javadetails/java-0542.html

Answer 6

Adding up on EJP's answer; use this for more fluidity. Make sure you don't put his code inside a bigger try catch with more code between the .read and the catch block, it may return an exception and jump all the way to the outer catch block, safest bet is to place EJPS's while loop inside a try catch, and then continue the code after it, like:

int count;
byte[] bytes = new byte[4096];
try {
    while ((count = is.read(bytes)) > 0) {
        System.out.println(count);
        bos.write(bytes, 0, count);
    }
} catch ( Exception e )
{
    //It will land here....
}
// Then continue from here

EDIT: ^This happened to me cuz I didn't realize you need to put socket.shutDownOutput() if it's a client-to-server stream!

Hope this post solves any of your issues