Generating Unique identifier using two strings

Question

I would like to get an idea on how can we generate an unique identifier using two Strings. My requirement here is to generate an unique identifier for a particular document. For id generation, document 'name' and 'version' has to be used. And there should be a way to get back both 'name' and 'version' from the unique identifier of a particular document. Is there a way to do this using UUID in java? or what is the best way of doing this. Can we use hashing or encoding for this purpose and if so how?

Answer 1

I don't know why you want to use 2 strings to generate a unique ID and it might not be possible to preserve uniqueness in some cases. java.util.UUID presents useful methods for your case. Take a look at this usage:

import java.util.UUID;

...

UUID idOne = UUID.randomUUID();
UUID idTwo = UUID.randomUUID();

If you are not satisfied with the IDs generated with this way, you can append / prepend your additional parameters like name and version to the generated IDs.

Answer 2

The best way to do this is to concatenate the strings with a separator which would not normally appear in those Strings

e.g.

String name = ....
String version = .....
String key = name + "/" + version;

You can obtain the original name and version with split("/")

Answer 3

You can use the static factory method randomUUID() to get a UUID object:

UUID id = UUID.randomUUID();    

To combine the id with the version, come up with a delimiter that you know won't be in either string, like / or _. Then you can split on that delimiter or use regular expressions to extract what you desire:

String entry = id + "_" + version;
String[] divided = entry.split(delimiter);

//or using regex

String entry= "idName_version"; //delimiter is "_"
String pattern = "(.*)_(.*)";

Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(example);

if (m.find())
{
    System.out.println(m.group(1)); //prints idName
    System.out.println(m.group(2)); //prints version
}

Answer 4

If you don't want to create a UUID just keep it pure strings

String.format("%s_%s_%s_%s", str1.length(), str2.length(), str1, str2);

tested with the following

Pair.of("a", ""),      // 'a'   and ''    into '1_0_a_'
Pair.of("", "a"),      // ''    and 'a'   into '0_1__a'
Pair.of("a", "a"),     // 'a'   and 'a'   into '1_1_a_a'
Pair.of("aa", "a"),    // 'aa'  and 'a'   into '2_1_aa_a'
Pair.of("a", "aa"),    // 'a'   and 'aa'  into '1_2_a_aa'
Pair.of("_", ""),      // '_'   and ''    into '1_0___'
Pair.of("", "_"),      // ''    and '_'   into '0_1___'
Pair.of("__", "_"),    // '__'  and '_'   into '2_1_____'
Pair.of("_", "__"),    // '_'   and '__'  into '1_2_____'
Pair.of("__", "__"),   // '__'  and '__'  into '2_2______'
Pair.of("/t/", "/t/"), // '/t/' and '/t/' into '3_3_/t/_/t/'
Pair.of("", "")        // ''    and ''    into '0_0__'

This works because the beginning creates a way to parse the following using a delimiter that cannot exist with numbers, it would fail if one used a delimiter like '0'

The other examples rely on the delimiter not existing in either string

str1+"."+str2
// both "..","." and ".",".." result in "...."

EDIT

Support for combining 'n' strings

private static String getUniqueString(String...srcs) {
  StringBuilder uniqueCombo = new StringBuilder();
  for (String src : srcs) {
    uniqueCombo.append(src.length());
    uniqueCombo.append('_');
  }
  // adding this second _ between numbers and strings allows unique strings across sizes
  uniqueCombo.append('_');
  for (String src : srcs) {
    uniqueCombo.append(src);
    uniqueCombo.append('_');
  }
  return uniqueCombo.toString();
}