Question

我正在使用的一些字符串模式并没有在运行时获取匹配，而当我在线使用正则表达式检查程序（如Regular Expression Tester时）时，我看到我的模式会获取所需的输出。

public class RegexMatches
{
    public static void main( String args[] ){
        String input = "[1463895254]PING www.andi.dz (213.179.181.44) 100(128) bytes of data.[1463895254]108 bytes from 213.179.181.44: icmp_seq=1 ttl=54 time=195 ms[1463895255]108 bytes from 213.179.181.44: icmp_seq=2 ttl=54 time=202 ms[1463895256]108 bytes from 213.179.181.44: icmp_seq=3 ttl=54 time=180 ms[1463895257]108 bytes from 213.179.181.44: icmp_seq=4 ttl=54 time=200 ms[1463895258]108 bytes from 213.179.181.44: icmp_seq=5 ttl=54 time=206 ms[1463895259]108 bytes from 213.179.181.44: icmp_seq=6 ttl=54 time=188 ms[1463895260]108 bytes from 213.179.181.44: icmp_seq=7 ttl=54 time=182 ms[1463895261]108 bytes from 213.179.181.44: icmp_seq=8 ttl=54 time=223 ms[1463895263]108 bytes from 213.179.181.44: icmp_seq=9 ttl=54 time=187 ms[1463895263]108 bytes from 213.179.181.44: icmp_seq=10 ttl=54 time=199 ms";

        try{
            String op[] = parseIndividualIP(input);
            System.out.println(op[1]);             
            System.out.println(op[8]);                 
        }
        catch (Exception e){
            e.printStackTrace();
        }
    }

// Doesn't Work...
    public static String[] parseIndividualIP(String input) throws IPAddressNotFoundException {
        // Capture the ip-address after 'x byes from'
        Pattern p = Pattern.compile("bytes\\s+from\\s+([\\d,\\.]+):");
        Matcher m = p.matcher(input);
        if (m.find()){
            int i = 0;
            String[] s = new  String[10];
            while(m.find()){
                s[i] = m.group(++i);
            }
            return s;
        }
        else
            throw new IPAddressNotFoundException();
    }

}

我不知道为什么在运行时我没有匹配以及我应该如何调试此问题。由于模式在运行之前和之后进行了交叉检查。

输入字符串 -

[1463895254]PING www.andi.dz (213.179.181.44) 100(128) bytes of data.[1463895254]108 bytes from 213.179.181.44: icmp_seq=1 ttl=54 time=195 ms[1463895255]108 bytes from 213.179.181.44: icmp_seq=2 ttl=54 time=202 ms[1463895256]108 bytes from 213.179.181.44: icmp_seq=3 ttl=54 time=180 ms[1463895257]108 bytes from 213.179.181.44: icmp_seq=4 ttl=54 time=200 ms[1463895258]108 bytes from 213.179.181.44: icmp_seq=5 ttl=54 time=206 ms[1463895259]108 bytes from 213.179.181.44: icmp_seq=6 ttl=54 time=188 ms[1463895260]108 bytes from 213.179.181.44: icmp_seq=7 ttl=54 time=182 ms[1463895261]108 bytes from 213.179.181.44: icmp_seq=8 ttl=54 time=223 ms[1463895263]108 bytes from 213.179.181.44: icmp_seq=9 ttl=54 time=187 ms[1463895263]108 bytes from 213.179.181.44: icmp_seq=10 ttl=54 time=199 ms

使用正则表达式模式 -

bytes\\s+from\\s+([\\d,\\.]+):

Answer 1

主要问题是使用m.group(++i)。在正则表达式中，您有一个捕获组来捕获IP地址（([\\d,\\.]+)），这意味着您应该调用m.group(1)，因为它返回第一组捕获的字符串。

Pattern对象是线程安全的，因此可以编译一次并重复使用相同的实例。

以下代码包含组修复以及一些可读性修改。将数组的用法更改为链接列表，而不是抛出异常，方法返回，如果找不到匹配项，则返回空列表。

import java.util.LinkedList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexMatches {
    private static final Pattern PATTERN = Pattern.compile("bytes\\s+from\\s+([\\d,\\.]+):");

    public static void main(String args[]) {
        String input = "[1463895254]PING www.andi.dz (213.179.181.44) 100(128) bytes of data.[1463895254]108 bytes from 213.179.181.44: icmp_seq=1 ttl=54 time=195 ms[1463895255]108 bytes from 213.179.181.44: icmp_seq=2 ttl=54 time=202 ms[1463895256]108 bytes from 213.179.181.44: icmp_seq=3 ttl=54 time=180 ms[1463895257]108 bytes from 213.179.181.44: icmp_seq=4 ttl=54 time=200 ms[1463895258]108 bytes from 213.179.181.44: icmp_seq=5 ttl=54 time=206 ms[1463895259]108 bytes from 213.179.181.44: icmp_seq=6 ttl=54 time=188 ms[1463895260]108 bytes from 213.179.181.44: icmp_seq=7 ttl=54 time=182 ms[1463895261]108 bytes from 213.179.181.44: icmp_seq=8 ttl=54 time=223 ms[1463895263]108 bytes from 213.179.181.44: icmp_seq=9 ttl=54 time=187 ms[1463895263]108 bytes from 213.179.181.44: icmp_seq=10 ttl=54 time=199 ms";

        List<String> op = parseIndividualIP(input);
        op.forEach(System.out::println);
    }

    public static List<String> parseIndividualIP(String input) {
        Matcher m = PATTERN.matcher(input);
        List<String> ips = new LinkedList<>();
        while (m.find()) {
            ips.add(m.group(1));
        }
        return ips;
    }
}

正则表达式模式不匹配

1 个答案: