在 Java 中验证 IPv4 地址

IT小君   2021-09-26T06:38:31

我想使用 Java 验证 IPv4 地址。它应该使用点十进制表示法编写,因此它应该有 3 个点 (" ."),没有字符,点之间有数字,并且数字应该在有效范围内。应该怎么做?

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评论(22)
IT小君

使用正则表达式非常简单(但请注意 ,与使用 Apache Commons Utility 的worpet 的答案相比这效率低得多且难以阅读

private static final Pattern PATTERN = Pattern.compile(
        "^(([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.){3}([01]?\\d\\d?|2[0-4]\\d|25[0-5])$");

public static boolean validate(final String ip) {
    return PATTERN.matcher(ip).matches();
}

基于后Mkyong

2021-09-26T06:38:31   回复
IT小君
2021-09-26T06:38:31   回复
IT小君

使用番石榴

InetAddresses.isInetAddress(ipStr)
2021-09-26T06:38:32   回复
IT小君

请参阅使用 apache 公共库的https://stackoverflow.com/a/5668971/1586965InetAddressValidator

或者你可以使用这个功能 -

public static boolean validate(final String ip) {
    String PATTERN = "^((0|1\\d?\\d?|2[0-4]?\\d?|25[0-5]?|[3-9]\\d?)\\.){3}(0|1\\d?\\d?|2[0-4]?\\d?|25[0-5]?|[3-9]\\d?)$";

    return ip.matches(PATTERN);
}
2021-09-26T06:38:32   回复
IT小君

您可以使用正则表达式,如下所示:

(([0-1]?[0-9]{1,2}\.)|(2[0-4][0-9]\.)|(25[0-5]\.)){3}(([0-1]?[0-9]{1,2})|(2[0-4][0-9])|(25[0-5]))

这个验证值在范围内。

Android 支持正则表达式。请参阅java.util.regex.Pattern

class ValidateIPV4
{

   static private final String IPV4_REGEX = "(([0-1]?[0-9]{1,2}\\.)|(2[0-4][0-9]\\.)|(25[0-5]\\.)){3}(([0-1]?[0-9]{1,2})|(2[0-4][0-9])|(25[0-5]))";
   static private Pattern IPV4_PATTERN = Pattern.compile(IPV4_REGEX);

   public static boolean isValidIPV4(final String s)
   {          
      return IPV4_PATTERN.matcher(s).matches();
   }
}

为了避免一遍又一遍地重新编译模式,最好进行Pattern.compile()调用,使其只执行一次。

2021-09-26T06:38:32   回复
IT小君

还有一个未记录的实用程序类sun.net.util.IPAddressUtil您实际上不应该使用它,尽管它可能在快速一次性使用的实用程序中很有用:

boolean isIP = IPAddressUtil.isIPv4LiteralAddress(ipAddressString);

在内部,这是InetAddress用于解析 IP 地址的实用程序类

请注意,对于像“123”这样的字符串,这将返回 true,这些字符串在技术上是有效的 IPv4 地址,只是不是点十进制表示法。

2021-09-26T06:38:32   回复
IT小君

IPAddress Java 库会做到这一点。链接中提供了 javadoc。免责声明:我是项目经理。

该库透明地支持 IPv4 和 IPv6,因此验证两者的工作方式如下相同,并且它还支持 CIDR 子网。

验证地址是否有效

    String str = "1.2.3.4";
    IPAddressString addrString = new IPAddressString(str);
    try {
         IPAddress addr = addrString.toAddress();
         ...
    } catch(AddressStringException e) {
        //e.getMessage provides validation issue
    }
2021-09-26T06:38:33   回复
IT小君

这适用于 Android,测试IPv4IPv6

注意:常用的InetAddressUtils已弃用。使用新InetAddress

public static Boolean isIPv4Address(String address) {
    if (address.isEmpty()) {
        return false;
    }
    try {
        Object res = InetAddress.getByName(address);
        return res instanceof Inet4Address || res instanceof Inet6Address;
    } catch (final UnknownHostException exception) {
        return false;
    }
}
2021-09-26T06:38:33   回复
IT小君

编写一个合适的正则表达式并根据它进行验证。JVM 完全支持正则表达式。

2021-09-26T06:38:33   回复
IT小君

如果是IP4,可以使用正则表达式如下:

^(2[0-5][0-5])|(1\\d\\d)|([1-9]?\\d)\\.){3}(2[0-5][0-5])|(1\\d\\d)|([1-9]?\\d)$.

2021-09-26T06:38:34   回复
IT小君

有很多方法可以实现,但正则表达式更有效。

看看下面的代码:

public static void main(String[] args) {

    String ipStr1 = "255.245.188.123"; // valid IP address
    String ipStr2 = "255.245.188.273"; // invalid IP address - 273 is greater than 255

    validateIP(ipStr1);
    validateIP(ipStr2);
}

public static void validateIP(String ipStr) {
    String regex = "\\b((25[0–5]|2[0–4]\\d|[01]?\\d\\d?)(\\.)){3}(25[0–5]|2[0–4]\\d|[01]?\\d\\d?)\\b";
    System.out.println(ipStr + " is valid? " + Pattern.matches(regex, ipStr));
}
2021-09-26T06:38:34   回复
IT小君

正则表达式是解决这个问题最有效的方法。看看下面的代码。除了有效性,它还检查它所属的IP地址类别以及它是否是保留的IP地址

Pattern ipPattern;
int[] arr=new int[4];
int i=0;

//Method to check validity
 private String validateIpAddress(String ipAddress) {
      Matcher ipMatcher=ipPattern.matcher(ipAddress);

        //Condition to check input IP format
        if(ipMatcher.matches()) {       

           //Split input IP Address on basis of .
           String[] octate=ipAddress.split("[.]");     
           for(String x:octate) { 

              //Convert String number into integer
              arr[i]=Integer.parseInt(x);             
              i++;
         }

        //Check whether input is Class A IP Address or not
         if(arr[0]<=127) {                          
             if(arr[0]==0||arr[0]==127)
                 return(" is Reserved IP Address of Class A");
             else if(arr[1]==0&&arr[2]==0&&arr[3]==0)
                 return(" is Class A Network address");
             else if(arr[1]==255&&arr[2]==255&&arr[3]==255)
                 return( " is Class A Broadcast address");
             else 
                 return(" is valid IP Address of Class A");
         }

        //Check whether input is Class B IP Address or not
         else if(arr[0]>=128&&arr[0]<=191) {        
             if(arr[2]==0&&arr[3]==0)
                 return(" is Class B Network address");
             else if(arr[2]==255&&arr[3]==255)
                 return(" is Class B Broadcast address");
             else
                 return(" is valid IP Address of Class B");
         }

        //Check whether input is Class C IP Address or not
         else if(arr[0]>=192&&arr[0]<=223) {        
             if(arr[3]==0)
                 return(" is Class C Network address");
             else if(arr[3]==255)
                 return(" is Class C Broadcast address");
             else
                 return( " is valid IP Address of Class C");
        }

        //Check whether input is Class D IP Address or not
        else if(arr[0]>=224&&arr[0]<=239) {          
             return(" is Class D IP Address Reserved for multicasting");
        }

        //Execute if input is Class E IP Address
        else  {                                   
             return(" is Class E IP Address Reserved for Research and Development by DOD");
        }

    }

    //Input not matched with IP Address pattern
    else                                     
        return(" is Invalid IP Address");


}


public static void main(String[] args) {

    Scanner scan= new Scanner(System.in);
    System.out.println("Enter IP Address: ");

    //Input IP Address from user
    String ipAddress=scan.nextLine();  
    scan.close();
    IPAddress obj=new IPAddress();

    //Regex for IP Address
    obj.ipPattern=Pattern.compile("((([0-1]?\\d\\d?|2[0-4]\\d|25[0-5])\\.){3}([0-1]?\\d\\d?|2[0-4]\\d|25[0-5]))");

    //Display output
    System.out.println(ipAddress+ obj.validateIpAddress(ipAddress));

}
2021-09-26T06:38:34   回复
IT小君

使用正则表达式在两行中获取有效的 ip 地址请检查代码的注释会话正则表达式如何工作以获取数字范围。

public class regexTest {


    public static void main(String[] args) {
        String IP = "255.001.001.255";
        System.out.println(IP.matches(new MyRegex().pattern));
    }

    }

    /*
    * /d - stands for any number between 0 to 9
    * /d{1,2} - preceding number that 0 to 9 here , can be of 1 digit to 2 digit . so minimum 0 and maximum 99
    * |  this stands for or operator
    *
    * () this is applied on a group to get the single value of outcome
    * (0|1)\d{2} = first digit is either 0 or 1 and other two digits can be any number between ( 0 to 9)
    * 2[0-4]\d - first digit is 2 , second digit can be between 0 to 4 and last digit can be 0 to 9
    * 25[0-5] - first two digit will be 25 and last digit will be between 0 to 5
    *
    * */
    class MyRegex {

        String zeroTo255 = "(\\d{1,2}|(0|1)\\d{2}|2[0-4]\\d|25[0-5])";
        public String pattern =  zeroTo255 + "\\." + zeroTo255 + "\\." + zeroTo255 + "\\." + zeroTo255;;

    }
2021-09-26T06:38:34   回复
IT小君

请查看sun.net.util 中存在的IPAddressUtil OOTB 类,这应该对您有所帮助。

2021-09-26T06:38:35   回复
IT小君

如果您不关心范围,以下表达式将有助于验证从 1.1.1.1 到 999.999.999.999

"[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}"
2021-09-26T06:38:35   回复
IT小君
public static boolean isIpv4(String ipAddress) {
    if (ipAddress == null) {
        return false;
    }
    String ip = "^(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|[1-9])\\."
            + "(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|\\d)\\."
            + "(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|\\d)\\."
            + "(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|\\d)$";
    Pattern pattern = Pattern.compile(ip);
    Matcher matcher = pattern.matcher(ipAddress);
    return matcher.matches();
}
2021-09-26T06:38:35   回复
IT小君
/**
 * Check if ip is valid
 *
 * @param ip to be checked
 * @return <tt>true</tt> if <tt>ip</tt> is valid, otherwise <tt>false</tt>
 */
private static boolean isValid(String ip) {
    String[] bits = ip.split("\\.");
    if (bits.length != 4) {
        return false;
    }
    for (String bit : bits) {
        try {
            if (Integer.valueOf(bit) < 0 || Integer.valueOf(bit) > 255) {
                return false;
            }
        } catch (NumberFormatException e) {
            return false; /* contains other other character */
        }
    }
    return true;
}
2021-09-26T06:38:35   回复
IT小君

apache-httpcomponents 的库

// ipv4 is true
assertTrue(InetAddressUtils.isIPv4Address("127.0.0.1"));
// not detect the ipv6
assertFalse(InetAddressUtils.isIPv4Address("2001:0db8:85a3:0000:0000:8a2e:0370:7334"));

maven lib(2019 年 9 月更新)

<!-- https://mvnrepository.com/artifact/org.apache.httpcomponents/httpclient -->
<dependency>
    <groupId>org.apache.httpcomponents</groupId>
    <artifactId>httpclient</artifactId>
    <version>4.5.10</version>
</dependency>
2021-09-26T06:38:36   回复
IT小君

我的解决方案(支持前导 0):

   String pattern="^[0-9](\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25?[0-5])?\\.(\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25[0-5])?\\.(\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25[0-5])?\\.(\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25[0-5])?$";
2021-09-26T06:38:36   回复
IT小君
 private static final String IPV4_PATTERN_ALLOW_LEADING_ZERO =
            "^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
            "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
            "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
            "([01]?\\d\\d?|2[0-4]\\d|25[0-5])$";
2021-09-26T06:38:36   回复
IT小君
private static boolean findValidIP(String ipAddress) {
        String[] ipAddressSplited = ipAddress.split("\\.");
        int correctFragments = 0;

        for (String ctr : ipAddressSplited) {
            int value = -10;
            try {
                value = Integer.parseInt(ctr);
            } catch (NumberFormatException exception) {
                value = -1;
            }

            if (value >= 0 && value <= 255 && ipAddressSplited.length == 4) correctFragments++;
        }

        System.out.println(correctFragments == 4 ? "Valid IP Address - " + ipAddress : "Invalid IP Address - " + ipAddress);
        return correctFragments == 4;
    }
2021-09-26T06:38:37   回复
IT小君

public void setIpAddress(String ipAddress) { if(ipAddress.matches("^(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9] [0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]? )$")) //ipv4的正则表达式 this.ipAddress = ipAddress; else System.out.println("IP 地址不正确"); }

2021-09-26T06:38:37   回复