我正在使用无连接套接字来创建远程字符串处理器。用户输入两个术语和一个整数，全部用空格分隔。此输入变为两个字符串和一个整数。如果输入了无效的整数，则会显示错误消息并结束程序。如果输入了有效整数，则对这两个字符串执行某种方法。在执行该方法之后，我想将结果与我添加的单词（参见案例1）一起保存到变量并将其发送给用户。这是应用程序的发送者 - 接收者部分。有一个接收器 - 发件人部分，我没有发布。我还没有完成案例2-5。一旦我找到了如何做案例1，也许我会得到足够的指导来自己完成这些。
import java.net.*;
import java.io.*;
class datagramSR{
public static void main(String[ ] args){
try{
InetAddress receiverHost = InetAddress.getByName(args[0]);
int receiverPort = Integer.parseInt(args[1]);
System.println("Input is accepted in the form of term1 term2 choice\n");
System.println("To lexicographically compare terms, choose 1");
System.println("To append term2 to term1, choose 2");
System.println("To determine if term1 ends with term2, choose 3");
System.println("To determine if term1 starts with term2, choose 4");
System.println("To return the index of the first occurrence of term2 in term1, choose 5");
DatagramSocket mySocket = new DatagramSocket(message);
//tokenize input into 2 strings
StringTokenizer stack = new StringTokenizer();
String string1 = stack.nextToken();
String string2 = stack.nextToken();
//cases for method codes
int code = Integer.parseInt(stack.nextToken());
if( code==1 || code==2 || code==3 || code==4 || code==5) {
switch(code) {
case 1: string1.compareTo( string2 );
if(>0) { message = string1" comes after "string2" lexicographically";
}
else if(<0) { message = string1" comes before "string2" lexicographically";
}
else if(0) { message = Term1 and term2 are the same;
}
byte[] sendBuffer = message.getBytes( );
DatagramPacket packet = new DatagramPacket(sendBuffer, sendBuffer.length,receiverHost, receiverPort);
mySocket.send(packet);
break;
//case 2: string1.concat(string2);
//DatagramPacket packet = new //DatagramPacket(sendBuffer, sendBuffer.length,receiverHost, receiverPort);
//mySocket.send(packet);
// break;
//case 3: string1.endsWith(string2);
// DatagramPacket packet = new DatagramPacket(sendBuffer, sendBuffer.length,receiverHost, receiverPort);
// mySocket.send(packet);
//break;
//case 4: string1.startsWith(string2);
//DatagramPacket packet = new DatagramPacket(sendBuffer, sendBuffer.length,receiverHost, receiverPort);
// mySocket.send(packet);
//break;
//case 5: string1.indexOf(string2);
//DatagramPacket packet = new DatagramPacket(sendBuffer, sendBuffer.length,receiverHost, receiverPort);
//mySocket.send(packet);
//break;
}
}
else
System.out.println("invalid string..");
System.exit();
}
// to receive a message
int MESSAGE_LEN = 60;
byte[ ] recvBuffer = new byte[MESSAGE_LEN];
DatagramPacket datagram = new DatagramPacket(recvBuffer,MESSAGE_LEN);
mySocket.receive(datagram);
String recvdString = new String(recvBuffer);
System.out.println(“\n”+recvdString);
mySocket.close( );
}
catch(Exception e){
e.printStackTrace( );
}
}
}
答案 0 :(得分：0)
有些人可能会告诉我，你的问题对解决的问题缺乏最小的理解，因为它充满了编译错误，但我很好奇测试套接字（我以前从未想过有时间），所以很好地解决了这个问题。您的代码并在本地进行了测试。
结果，把它分解成一个可重复使用的类，而不是一个长的主方法，所以请耐心等待。
观察：
类以Java命名标准以大写字母开头。此外，Java中的常量通常定义为static final字段。
public class DatagramSR {
private static final int MESSAGE_LEN = 60;
private DatagramSocket socket;
private InetAddress host;
private int port;
public DatagramSR(InetAddress host, int port) throws SocketException {
this.socket = new DatagramSocket(port, host);
this.host = host;
this.port = port;
}
[...]
}
执行正确的错误处理，而不仅仅是捕获基本异常，然后在您的应用程序不起作用时想知道出了什么问题：
public static void main(String[] args) {
DatagramSR sr = null;
try {
InetAddress receiverHost = InetAddress.getByName(args[0]);
int receiverPort = Integer.parseInt(args[1]);
DatagramSocket mySocket = new DatagramSocket(receiverPort, receiverHost);
sr = new DatagramSR(mySocket, receiverHost, receiverPort);
} catch (ArrayIndexOutOfBoundsException e) {
System.out.println("Not enough parameters given.");
} catch (UnknownHostException e) {
System.out.println("Give a reasonable hostname as first parameter.");
} catch (NumberFormatException e) {
System.out.println("Give a reasonable port number as second parameter.");
} catch (SocketException e) {
System.out.println("Socket creation failed.");
}
// If it's still null, there were errors above and no point to proceed.
if (sr != null) {
sr.run();
}
}
例如，使用Scanner来获取用户输入。
Scanner scanner = new Scanner(System.in);
System.out.println("Define term1.");
String term1 = scanner.nextLine();
System.out.println("Define term2.");
String term2 = scanner.nextLine();
System.out.println("To lexicographically compare terms, choose 1");
System.out.println("To append term2 to term1, choose 2");
System.out.println("To determine if term1 ends with term2, choose 3");
System.out.println("To determine if term1 starts with term2, choose 4");
System.out.println("To return the index of the first occurrence of term2 in term1, choose 5");
while (true) {
try {
code = Integer.parseInt(scanner.nextLine());
break;
} catch (NumberFormatException e) {
System.out.println("Please give an integer to proceed.");
}
}
scanner.close();
如果代码在开关范围内，请不要特别检查。如果加上说几十个案例，那将只会成为一个野兽if条款。更容易抛出异常并在调用代码中处理它。也有助于干掉（不要重复自己）你的开关代码，然后为所有情况做一些常见的事情。
private void handleInput(String term1, String term2, int code)
throws IOException {
String message;
switch (code) {
case 1:
int comparison = term1.compareTo(term2);
if (comparison > 0) {
message = term1 + " comes after " + term2 + " lexicographically";
} else if (comparison < 0) {
message = term1 + " comes before " + term2 + " lexicographically";
} else {
message = "Term1 and term2 are the same";
}
break;
// Other cases omitted, since you said you could figure them out.
default:
throw new IOException("Unknown code: " + code);
}
byte[] sendBuffer = message.getBytes();
DatagramPacket packet = new DatagramPacket(sendBuffer, sendBuffer.length,
host, port);
socket.send(packet);
}