Where is the error in the server code?

Question

Good afternoon, tell me, I wrote a simple application, the client and the server, the task was to transfer files

CUSTOMER

public class Client_2 implements Runnable { static private ServerSocket server;//принимает соединение от Socket connection static private Socket client;//создаёт соединение ServerSocket так же может соедениться с channel socket /*****************STREAM*****************/ static private BufferedOutputStream output; static private BufferedInputStream input; /***************************************/ public static void main(String[] args) { new Thread (new Client_2()).start(); new Thread (new Server_2()).start(); } @Override public void run() { // TODO Auto-generated method stub byte[] byteArray = new byte[8192]; int in; try { client = new Socket("127.0.0.1", 60000); input = new BufferedInputStream(new FileInputStream("C:/tmp.txt")); output = new BufferedOutputStream(client.getOutputStream()); while ((in = input.read(byteArray)) != -1){ output.write(byteArray,0,in); } } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } try { input.close(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } try { output.close(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } }

SERVER

 public class Server_2 implements Runnable { static private ServerSocket server;//принимает соединение от Socket connection static private Socket client;//создаёт соединение ServerSocket так же может соедениться с channel socket /*****************STREAM*****************/ static private BufferedOutputStream output; static private BufferedInputStream input; /***************************************/ @Override public void run() { // TODO Auto-generated method stub byte[] byteArray = new byte[8192]; int in; try { server= new ServerSocket(60000/*номер порта*/, 10/*колличество подключений*/); output = new BufferedOutputStream(new FileOutputStream("D://")); input = new BufferedInputStream(client.getInputStream()); while ((in = ((ObjectInput) output).read(byteArray)) != -1){ ((ObjectOutput) input).write(byteArray,0,in); } } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } try { input.close(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } try { output.close(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } } }

but when you transfer the file gives an error

java.io.FileNotFoundException: D: \ (The system cannot find the specified path) at java.io.FileOutputStream.open0 (Native Method) at java.io.FileOutputStream.open (Unknown Source) at java.io.FileOutputStream. (Unknown Source) at java.io.FileOutputStream. (Unknown Source) at ru.cod.Server_2.run (Server_2.java:37) at java.lang.Thread.run (Unknown Source) Exception in thread "Thread-1" java. lang.NullPointerException at ru.cod.Server_2.run (Server_2.java:50) at java.lang.Thread.run (Unknown Source)

What could be a mistake?

Wrong way ..... and specifically new FileOutputStream("D://"));
The path "D://" cannot be a file, so FileOutputStream cannot open it for writing
In general, the server code is not working, starting with a client which nothing has been assigned, attempts to read from the output , to bring the BufferedInputStream to ObjectOutput .
And if it is not there, then create a file with the desired name. You don’t transfer the physical file itself, but only the content, so the D:// indication is incorrect
@zRrr Yes, the client is really empty, but what should it be assigned to?

Chad Chad 8,652 one 13 37 · Accepted Answer · 2016-06-10T06:00:18

You must specify the name of the file, in fact, as he swears.

 output = new BufferedOutputStream(new FileOutputStream("D://your_file_name"));

You are trying to write data to a directory. Of course, this is also a file, but it is special, it does not support the recording. Specify the required file name (or some temporary, or let the client additionally pass the file name)

Chad

8,652 one 13 37

2
Read the book on java. Any) - Stranger in the Q
one
You need a file name anyway. If you just want the data from the file to be stored - just create a temporary file (for example, the system File.createTempFile) or generate a unique name in the folder and save there. In any case, you can not save a file without a name. What is your task in the end that such strange thoughts on implementation arise? - Chad
one
If you need the file name to match the original, you need to transfer from the client past the file itself and its name. - Chad
one
Well, the most clumsy way, without bringing in new knowledge, is that you form a new file in which you first add its name. and complete the name \ x00 for example. And on the server from the stream, first isolate the name, and write the rest to the file. You can not create a new file on the client, but simply shove the file name first \ x00 into the output and then the file itself - Chad
one
those. you first output.write (fileNameBuf, 0, n) and therefore the file is already - Chad

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Where is the error in the server code?

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