Java ip2long equivalent

基本上，这会将您的点 IP 地址字符串转换为 long。

public static Long Dot2LongIP(String dottedIP) {
    String[] addrArray = dottedIP.split("''.");        
    long num = 0;        
    for (int i=0;i<addrArray.length;i++) {            
        int power = 3-i;            
        num += ((Integer.parseInt(addrArray[i]) % 256) * Math.pow(256,power));        
    }        
    return num;    
}

public static Long ipToLong(String stringIp) {
        return Arrays.stream(stringIp.split("''."))
                .mapToLong(Long::parseLong)
                .reduce(0,
                        (a, b) -> (a << 8) + b);
}
public static String ipToString(Long longIp){
        return ((longIp >> 24) & 0xFF) + "." +
                ((longIp >> 16) & 0xFF) + "." +
                ((longIp >> 8) & 0xFF) + "." +
                (longIp & 0xFF);
}

Java 中没有标准的 API 可以做到这一点，但是

1/InetAddress 类为您提供了一种获取字节数组的方法。

2/如果你真的需要一个整数，你可以使用这个片段，在 http://www.myteneo.net/blog/-/blogs/java-ip-address-to-integer-and-back/

public static String intToIp(int i) {
    return ((i >> 24 ) & 0xFF) + "." +
           ((i >> 16 ) & 0xFF) + "." +
           ((i >>  8 ) & 0xFF) + "." +
           ( i        & 0xFF);
}
public static Long ipToInt(String addr) {
    String[] addrArray = addr.split("''.");
    long num = 0;
    for (int i=0;i<addrArray.length;i++) {
        int power = 3-i;
        num += ((Integer.parseInt(addrArray[i])%256 * Math.pow(256,power)));
    }
    return num;
}

我将首先将字符串转换为八位字节：

static final String DEC_IPV4_PATTERN = "^(([0-1]?''d{1,2}''.)|(2[0-4]''d''.)|(25[0-5]''.)){3}(([0-1]?''d{1,2})|(2[0-4]''d)|(25[0-5]))$";
static byte[] toOctets(String address){
    if(address==null){
        throw new NullPointerException("The IPv4 address cannot be null.");
    }
    if(!address.matches(DEC_IPV4_PATTERN)){
        throw new IllegalArgumentException(String.format("The IPv4 address is invalid:%s ",address));
    }
    //separate octets into individual strings
    String[] numbers = address.split("''.");
    //convert octets to bytes.
    byte[] octets = new byte[4];
    for(int i = 0; i < octets.length; i++){
        octets[i] = Integer.valueOf(numbers[i]).byteValue();
    }
    return octets;
}

然后是八位字节到 BigInteger，因为它接受一个字节数组，并从它到一个整数：

static int toInteger(byte[] octets){
    if(octets==null){
        throw new NullPointerException("The array of octets cannot be null");
    }
    if(octets.length != 4){
        throw new IllegalArgumentException(String.format("The byte array must contain 4 octets: %d",octets.length));
    }
    return new BigInteger(octets).intValue();
}

从这里，您可以简单地执行以下操作：

String address = "127.0.0.1";
System.out.println(toInteger(toOctets(address)));

或者创建一个名为 ip2long(String address ) 的函数

这个适用于IPv4和IPv6：

public static String toLongString(final String ip) {
    try {
        final InetAddress address = InetAddress.getByName(ip);
        final byte[] bytes = address.getAddress();
        final byte[] uBytes = new byte[bytes.length + 1]; // add one byte at the top to make it unsigned
        System.arraycopy(bytes, 0, uBytes, 1, bytes.length);
        return new BigInteger(uBytes).toString();
    } catch (final UnknownHostException e) {
        throw new IllegalArgumentException(e);
    }
}

我希望下面的链接对您有所帮助，反之亦然

https://www.mkyong.com/java/java-convert-ip-address-to-decimal-number。

从上面的链接中提供的方法：

public static long ipToLong(String ipAddress) {
    String[] ipAddressInArray = ipAddress.split("''.");
    long result = 0;
    for (int i = 0; i < ipAddressInArray.length; i++) {     
        int power = 3 - i;
        int ip = Integer.parseInt(ipAddressInArray[i]);
        result += ip * Math.pow(256, power);
    }
    return result;
}
public static String longToIp(long ip) {
    StringBuilder result = new StringBuilder(15);
    for (int i = 0; i < 4; i++) {
        result.insert(0,Long.toString(ip & 0xff));
        if (i < 3) {
            result.insert(0,'.');
        }
        ip = ip >> 8;
    }
    return result.toString();
}