Java ip2long equivalent

Question

Hello I'm having a database to select the IP location from>

The script was in php and I'm converting it to java but I have no idea what is the equivalent of ip2long('127.0.0.1' )); in java

The basic formula for this can be found many places. [Here's one](http://guyellisrocks.com/coding/converting-a-string-ip-to-an-int-and-back/) — Michael Berkowski, May 29 '12 at 14:06
possible duplicate of [Going from 127.0.0.1 to 2130706433, and back again](http://stackoverflow.com/questions/2241229/going-from-127-0-0-1-to-2130706433-and-back-again) — Michael Berkowski, May 29 '12 at 14:07

score 3 · Accepted Answer · answered May 29 '12 at 14:06

3

Basically, this will convert your dotted IP address string to 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;    
}

answered May 29 '12 at 14:06

Siva Charan

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thanks alot will have to try it to make sure it gets me the exact results. you saved me time :) – Lamis May 29 '12 at 14:09

score 2 · Answer 2 · answered Apr 08 '20 at 14:12

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);
}

score 1 · Answer 3 · answered May 29 '12 at 14:10

I don't think there is a standard API to do that in Java, but

1/ The InetAddress class gives you a method to get an array of byte.

2/ If you really need a single integer, you can use this snippet, found on 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;

}

Don't try that "too" actually it looks like we have the same source, except the fact that I quoted mine ;=) — Samuel Rossille, May 29 '12 at 14:13

score 1 · Answer 4 · answered May 29 '12 at 14:20

I would start by converting the string to octets:

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;
}

And then the octets to a BigInteger since it accepts a byte array, and from it to an integer:

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();
}

And from here, you can simply do:

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

Or create a function named ip2long(String address )

score 0 · Answer 5 · answered Nov 03 '16 at 15:57

This one works for IPv4 and 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);
    }
}

Anil kumar · Answer 6 · 2018-11-02T06:17:52.930

I hope below link will help you, clearly provided vice versa too

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

Provided methods from above link:

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();
}

Java ip2long equivalent

6 Answers6