java网络编程应该怎样在客户端和服务器间实现通信？

以前写的，照贴了。。。服务器端：import java.awt.*\x0d\x0aimport java.awt.event.WindowAdapter\x0d\x0aimport java.awt.event.WindowEvent\x0d\x0aimport java.io.*\x0d\x0aimport java.net.*/*6、 采用UDP协议，编写一个Java网络应用程序，该应用分服务器端程序和客户端程序两部分。\x0d\x0a* 客户端指定一个服务器上的文件名，让服务器发回该文件的内容，或者提示文件不存在。\x0d\x0a* （20分）（服务端程序和客户端程序分别命名为Server.java和Client.java）*/\x0d\x0apublic class N4BT6 extends Frame\x0d\x0a{\x0d\x0aDatagramSocket socket \x0d\x0aDatagramPacket packet byte[] buf \x0d\x0aFile file \x0d\x0aFileInputStream input\x0d\x0aString message = "该文件不存在"\x0d\x0aTextArea text\x0d\x0apublic N4BT6(String title)\x0d\x0a{\x0d\x0asuper(title)\x0d\x0atext = new TextArea(6,4)\x0d\x0aadd(text)\x0d\x0asetSize(400, 300)\x0d\x0asetVisible(true)\x0d\x0aaddWindowListener(new WindowAdapter()\x0d\x0a{\x0d\x0apublic void windowClosing(WindowEvent e)\x0d\x0a{\x0d\x0adispose()\x0d\x0a}\x0d\x0a})\x0d\x0a\x0d\x0abuf = new byte[1024]\x0d\x0atry\x0d\x0a{\x0d\x0asocket = new DatagramSocket(1230)\x0d\x0apacket = new DatagramPacket(buf, buf.length)\x0d\x0asocket.receive(packet)\x0d\x0afile = new File(new String(packet.getData()))\x0d\x0asocket = new DatagramSocket()\x0d\x0a} \x0d\x0acatch (Exception e)\x0d\x0a{e.printStackTrace()\x0d\x0a}\x0d\x0a\x0d\x0aif(file.exists())\x0d\x0a{\x0d\x0atry\x0d\x0a{\x0d\x0abuf = new byte[(int)file.length()]\x0d\x0apacket = new DatagramPacket(buf,buf.length,InetAddress.getLocalHost(),1234)\x0d\x0ainput = new FileInputStream(file)\x0d\x0ainput.read(buf)\x0d\x0asocket.send(packet)\x0d\x0a}\x0d\x0acatch (IOException e) \x0d\x0a{\x0d\x0ae.printStackTrace()\x0d\x0a}\x0d\x0a}\x0d\x0aelse\x0d\x0a{\x0d\x0atry\x0d\x0a{\x0d\x0apacket = new DatagramPacket(message.getBytes(),message.getBytes().length,\x0d\x0aInetAddress.getLocalHost(),1234)\x0d\x0asocket.send(packet)\x0d\x0a}\x0d\x0acatch (Exception e) \x0d\x0a{\x0d\x0ae.printStackTrace()\x0d\x0a}\x0d\x0a}\x0d\x0a\x0d\x0a}\x0d\x0apublic static void main(String[] args)\x0d\x0a{\x0d\x0anew N4BT6("Server")\x0d\x0a}\x0d\x0a}\x0d\x0a客户端：import java.awt.*\x0d\x0aimport java.awt.event.*\x0d\x0aimport java.net.DatagramPacket\x0d\x0aimport java.net.DatagramSocket\x0d\x0aimport java.net.InetAddresspublic class N4BT6_2 extends Frame\x0d\x0a{\x0d\x0aTextArea text\x0d\x0aString message = "Q.txt"\x0d\x0aDatagramSocket socket \x0d\x0aDatagramPacket packet\x0d\x0abyte[] buf\x0d\x0apublic N4BT6_2(String title)\x0d\x0a{\x0d\x0asuper(title)\x0d\x0atext = new TextArea(6,4)\x0d\x0aadd(text)\x0d\x0asetSize(400, 300)\x0d\x0asetVisible(true)\x0d\x0aaddWindowListener(new WindowAdapter()\x0d\x0a{\x0d\x0apublic void windowClosing(WindowEvent e)\x0d\x0a{\x0d\x0adispose()\x0d\x0a}\x0d\x0a})\x0d\x0atry\x0d\x0a{\x0d\x0a\x0d\x0asocket = new DatagramSocket()\x0d\x0apacket = new DatagramPacket(message.getBytes(),message.getBytes().length,\x0d\x0aInetAddress.getLocalHost(),1230)\x0d\x0asocket.send(packet)\x0d\x0a}\x0d\x0acatch (Exception e) \x0d\x0a{\x0d\x0ae.printStackTrace()\x0d\x0a}\x0d\x0a\x0d\x0atry\x0d\x0a{\x0d\x0abuf = new byte[1024]\x0d\x0asocket = new DatagramSocket(1234)\x0d\x0apacket = new DatagramPacket(buf,buf.length)\x0d\x0asocket.receive(packet)\x0d\x0atext.append(new String(buf))\x0d\x0a}\x0d\x0acatch (Exception e) \x0d\x0a{\x0d\x0ae.printStackTrace()\x0d\x0a}\x0d\x0a}\x0d\x0apublic static void main(String[] args)\x0d\x0a{\x0d\x0anew N4BT6_2("Client")\x0d\x0a}\x0d\x0a}

没有描述清楚。按照字面理解，服务端根据客户端发送的信息判断是否发送文件是吗？服务器端根据客户端请求的信息，如果要返回文件，服务器端读取文件向客户端写文件，比如读取文件向客户端写字节流。客户端肯定是不能直接访问服务器端文件的。如果文件是放在可以公开访问的地址，返回地址，客户端下载就是了。所以你的问题没有描述清楚。

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