获取java中“9.1.0.0-10”形式的I​​pAddresses范围

Question

需要获得9.1.0.0-10格式中提供的所有Ips

输出：

9.1.0.0
9.1.0.1
9.1.0.2
9.1.0.3
9.1.0.4
9.1.0.5
9.1.0.6
9.1.0.7
9.1.0.8
9.1.0.9
9.1.0.10

是否有任何课程可以完成这项工作，或者有任何方法可以更好地完成这项工作

我的java代码：

String start = "9.1.0.0-10";

         String range[] = start.split("-");
         String[] startParts = range[0].split("(?<=\\.)(?!.*\\.)");
         int first = Integer.parseInt(startParts[1]);
         int last = Integer.parseInt(range[1]);

         for(int i=first;i<=last;i++)
         System.out.println(startParts[0] + i);

Answer 1

首先，我建议您先按时间段分割地址范围，然后根据短划线的存在分别处理每个令牌。这样，你也可以处理像9.1.0-5.0-10这样的范围。

前段时间我不得不解决这个问题，这导致我创建了一个InetAddressRange类：

import java.net.InetAddress;
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

/**
 * Notation: n.n.n.n, where n can be:
 * - an integer such that 0 <= n <= 255;
 * - an interval m-n, where 0 <= m <= n <= 255
 * - an asterisk (*), synonym for interval 0-255.
 * The address parts are stored as returned as 16-bits integers (short) in order to
 * avoid confusion that might come from the fact that integers between 128 and 255
 * are considered negative (-128 to -1) when handled as byte values.
 * Instances of this class are immutable.
 */
public class InetAddressRange {

  private static final Pattern P_DIGITS = Pattern.compile("\\d+");
  private static final Pattern P1 = Pattern.compile("([0-9\\*\\-\\[\\],]+)\\.([0-9\\*\\-\\[\\],]+)\\.([0-9\\*\\-\\[\\],]+)\\.([0-9\\*\\-\\[\\],]+)");
  private static final Pattern P2 = Pattern.compile("(\\d+)\\-(\\d+)");

  private final short[] min = new short[4];
  private final short[] max = new short[4];

  public InetAddressRange(String expr) {
    parse(expr);
  }

  private void parse(String expr) {
    Matcher m1 = P1.matcher(expr);
    if (!m1.matches()) {
      throw error(expr);
    }

    for (int i = 0; i < 4; i++) {
      String token = m1.group(i + 1);
      if (P_DIGITS.matcher(token).matches()) {
        // Integer between 0 and 255 (else only the last 8 bits are kept)
        min[i] = max[i] = (short) (Integer.parseInt(token) & 0xff);
      } else if (token.equals("*")) {
        // Asterisk = interval 0-255
        min[i] = 0;
        max[i] = 0xff;
      } else {
        // Interval a-b
        Matcher m2 = P2.matcher(token);
        if (m2.matches()) {
          min[i] = (short) (Integer.parseInt(m2.group(1)) & 0xff);
          max[i] = (short) (Integer.parseInt(m2.group(2)) & 0xff);
        } else {
          throw error(expr);
        }
      }
    }
  }

  private static IllegalArgumentException error(String expr) {
    return new IllegalArgumentException("The expression " + expr + " isn't a valid IP address interval.");
  }

  public short getMin(int i) {
    return min[i];
  }

  public short getMax(int i) {
    return max[i];
  }

  public boolean isInRange(InetAddress adr) {
    byte[] parts = adr.getAddress();
    for (int i = 0; i < 4; i++) {
      if (min[i] > (parts[i] & 0xff) || max[i] < (parts[i] & 0xff)) {
        return false;
      }
    }
    return true;
  }

  public boolean isInRange(String adr) {
    String[] tokens = adr.split("\\.");
    if (tokens.length < 4) {
      return false;
    }
    for (int i = 0; i < 4; i++) {
      int part = Integer.parseInt(tokens[i]);
      if (min[i] > part || max[i] < part) {
        return false;
      }
    }
    return true;
  }

  public List<String> getAddresses() {
    List<String> addrs = new ArrayList<>();
    for (short a = min[0]; a <= max[0]; a++) {
      for (short b = min[1]; b <= max[1]; b++) {
        for (short c = min[2]; c <= max[2]; c++) {
          for (short d = min[3]; d <= max[3]; d++) {
            String addr = String.format("%d.%d.%d.%d", a, b, c, d);
            addrs.add(addr);
          }
        }
      }
    }
    return addrs;
  }

  public static void main(String[] args) {
    InetAddressRange range = new InetAddressRange("9.1.0.0-10");
    range.getAddresses().forEach(System.out::println);
  }

}

getAddresses()方法将生成您需要的地址（有关示例，请参阅main()方法）。

干杯，

杰夫