Java: Common way to validate and convert "host:port" to InetSocketAddress?

Question

What is the common way in Java to validate and convert a string of the form host:port into an instance of InetSocketAddress?

It would be nice if following criteria were met:

• No address lookups;

• Working for IPv4, IPv6, and "string" hostnames;
  (For IPv4 it's ip:port, for IPv6 it's [ip]:port, right? Is there some RFC which defines all these schemes?)

• Preferable without parsing the string by hand.
  (I'm thinking about all those special cases, when someone think he knows all valid forms of socket addresses, but forgets about "that special case" which leads to unexpected results.)

Answer 1

I myself propose one possible workaround solution.

Convert a string into URI (this would validate it automatically) and then query the URI's host and port components.

Sadly, an URI with a host component MUST have a scheme. This is why this solution is "not perfect".

String string = ... // some string which has to be validated

try {
  // WORKAROUND: add any scheme to make the resulting URI valid.
  URI uri = new URI("my://" + string); // may throw URISyntaxException
  String host = uri.getHost();
  int port = uri.getPort();

  if (uri.getHost() == null || uri.getPort() == -1) {
    throw new URISyntaxException(uri.toString(),
      "URI must have host and port parts");
  }

  // here, additional checks can be performed, such as
  // presence of path, query, fragment, ...


  // validation succeeded
  return new InetSocketAddress (host, port);

} catch (URISyntaxException ex) {
  // validation failed
}

This solution needs no custom string parsing, works with IPv4 (1.1.1.1:123), IPv6 ([::0]:123) and host names (my.host.com:123).

Accidentally, this solution is well suited for my scenario. I was going to use URI schemes anyway.

Answer 2

A regex will do this quite neatly:

Pattern p = Pattern.compile("^\\s*(.*?):(\\d+)\\s*$");
Matcher m = p.matcher("127.0.0.1:8080");
if (m.matches()) {
  String host = m.group(1);
  int port = Integer.parseInt(m.group(2));
}

You can this in many ways such as making the port optional or doing some validation on the host.

Answer 3

It doesn't answer the question exactly, but this answer could still be useful others like me who just want to parse a host and port, but not necessarily a full InetAddress. Guava has a HostAndPort class with a parseString method.

Answer 4

Another person has given a regex answer which is what I was doing to do when originally asking the question about hosts. I will still do because it's an example of a regex that is slightly more advanced and can help determine what kind of address you are dealing with.

String ipPattern = "(\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}\\.\\d{1,3}):(\\d+)";
String ipV6Pattern = "\\[([a-zA-Z0-9:]+)\\]:(\\d+)";
String hostPattern = "([\\w\\.\\-]+):(\\d+)";  // note will allow _ in host name
Pattern p = Pattern.compile( ipPattern + "|" + ipV6Pattern + "|" + hostPattern );
Matcher m = p.matcher( someString );
if( m.matches() ) {
    if( m.group(1) != null ) {
        // group(1) IP address, group(2) is port
    } else if( m.group(3) != null ) {
        // group(3) is IPv6 address, group(4) is port            
    } else if( m.group(5) != null ) {
        // group(5) is hostname, group(6) is port
    } else {
        // Not a valid address        
    }
}

Modifying so that port is optional is pretty straight forward. Wrap the ":(\d+)" as "(?::(\d+))?" and then check for null for group(2), etc.

Edit: I'll note that there's no "common way" way that I'm aware of but the above is how I'd do it if I had to.

Also note: the IPv4 case can be removed if the host and IPv4 cases will actually be handled the same. I split them out because sometimes you can avoid an ultimate host look-up if you know you have the IP address.

Answer 5

new InetSocketAddress(
  addressString.substring(0, addressString.lastIndexOf(":")),
  Integer.parseInt(addressString.substring(addressString.lastIndexOf(":")+1, addressString.length));

? I probably made some little silly mistake. and I'm assuming you just wanted a new InetSocketAddress object out of the String in only that format. host:port

Answer 6

All kind of peculiar hackery, and elegant but unsafe solutions provided elsewhere. Sometimes the inelegant brute-force solution is the way.

public static InetSocketAddress parseInetSocketAddress(String addressAndPort) throws IllegalArgumentException {
    int portPosition = addressAndPort.length();
    int portNumber = 0;
    while (portPosition > 1 && Character.isDigit(addressAndPort.charAt(portPosition-1)))
    {
        --portPosition;
    }
    String address;
    if (portPosition > 1 && addressAndPort.charAt(portPosition-1) == ':')
    {
        try {
            portNumber = Integer.parseInt(addressAndPort.substring(portPosition));
        } catch (NumberFormatException ignored)
        {
            throw new IllegalArgumentException("Invalid port number.");
        }
        address = addressAndPort.substring(0,portPosition-1);
    } else {
        portNumber = 0;
        address = addressAndPort;
    }
    return new InetSocketAddress(address,portNumber);
}

Answer 7

The open-source IPAddress Java library has a HostName class which will do the required parsing. Disclaimer: I am the project manager of the IPAddress library.

It will parse IPv4, IPv6 and string host names with or without ports. It will handle all the various formats of hosts and addresses. BTW, there is no single RFC for this, there are a number of RFCs that apply in different ways.

String hostName = "[a:b:c:d:e:f:a:b]:8080";
String hostName2 = "1.2.3.4:8080";
String hostName3 = "a.com:8080";
try {
    HostName host = new HostName(hostName);
    host.validate();
    InetSocketAddress address = host.asInetSocketAddress();
    HostName host2 = new HostName(hostName2);
    host2.validate();
    InetSocketAddress address2 = host2.asInetSocketAddress();
    HostName host3 = new HostName(hostName3);
    host3.validate();
    InetSocketAddress address3 = host3.asInetSocketAddress();
    // use socket address      
} catch (HostNameException e) {
    String msg = e.getMessage();
    // handle improperly formatted host name or address string
}

Answer 8

URI can accomplish this:

URI uri = new URI(null, "example.com:80", null, null, null);

Unfortunately, there's a bug in current OpenJDK (or in the documentation) where the authority isn't properly validated. The documentation states:

The resulting URI string is then parsed as if by invoking the URI(String) constructor and then invoking the parseServerAuthority() method upon the result

That call to parseServerAuthority just doesn't happen unfortunately so the real solution here that properly validates is:

URI uri = new URI(null, "example.com:80", null, null, null).parseServerAuthority();

then

InetSocketAddress address = new InetSocketAddress(uri.getHost(), uri.getPort());