java - 如何确保使用java.net传递的文件与传输前一样

Question

我正在研究java.net并尝试传输som文件。这是发件人代码

public static final FilesGetter filesGetter = new FilesGetter();
    public static Socket s;
    public static File[] files;

    public static void main(String args[]) throws Exception{
        s = new Socket("localhost", 3128);

        while (true){
            try{
                files = filesGetter.getFilesList("/etc/dlp/templates/");
                Socket s = new Socket("localhost", 3128);
                args[0] = args[0]+"\n"+s.getInetAddress().getHostAddress()
                        +":"+s.getLocalPort();
                if (files != null){
                    for (int i = 0; i < files.length; i++){
                        InputStream is = new FileInputStream(files[i]);
                        byte[] message = IOUtils.toByteArray(is);
                        s.getOutputStream().write(message);

                        byte buf[] = new byte[64*1024];
                        int r = s.getInputStream().read(buf);
                        String data = new String(buf, 0, r);

                        System.out.println(data);
                    }
                }
            } catch(Exception e){
                System.out.println("init error: "+e);
            }
        }    
    }

这是接收者代码:

public class Consumer extends Thread{

    public static Socket s;
    public String customerId;
    int num;
    public static final Filter filter = new Filter();
    public MimeParser mimeParser = new MimeParser(true);

    public Consumer(int num, final Socket s, final String customerId){
        this.num = num;
        this.s = s;
        this.customerId = customerId;

        setDaemon(true);
        setPriority(NORM_PRIORITY);
        start();
    }

    public static void receive(final String customerId){
        try {
            int i = 0;

            ServerSocket server = new ServerSocket(3128, 0, InetAddress.getByName("localhost"));

            System.out.println("server started");
            while (true){
                new Consumer(i, server.accept(), customerId);
                i++;
            }
        } catch (Exception e){
            e.printStackTrace();
        }
    }

    public void run(){
        try {
            InputStream is = s.getInputStream();
            OutputStream os = s.getOutputStream();

            byte buf[] = new byte[64*1024];
            int r = is.read(buf);

            if (r < 0)
                    return;
            ByteArrayInputStream bais = new ByteArrayInputStream(buf, 0, r);
            MessageInfo mi = mimeParser.parseMessages(bais);
            filter.getMessageAcceptability(mi.getPlainText(), customerId);
            s.close();
        } catch (Exception e){
            System.out.println("init error: " + e);
        }
    }
}

我不确定数据完整性，因为服务器端的数据处理并未完全成功，并且不知道是否需要查找处理代码中的错误(当我使用rabbitmq)或在客户端-服务器代码中。我也不知道必须选择什么缓冲区大小。

Answer 1

在发送端你可以发送任意数量的数据，64*1024就可以了，send方法会循环直到所有数据都发送完毕。但在接收方，read 可能会在整个文件被读取之前返回，您必须循环读取，直到另一端关闭套接字。

在这些情况下，最好提前发送一个整数，指示您要发送的数据量，接收器将循环直到读取那么多字节。

例如:

int ret=0;
int offset=0;
int BUFF_LEN=64*1024;
byte[] buffer = new byte[BUFF_LEN];
while ((ret = is.read(buffer, offset, BUFF_LEN - offset)) > 0)
{
    offset+=ret;
    // just in case the file is bigger that the buffer size
   if (offset >= BUFF_LEN) break;
}