在 Java 128 位 hash 中，32 位（13 位）将是纪元，其余位（86 位）一些 Z0800FC577294C34E0B28AD283943594

Question

In Java 128bit hash where 32bit (13 digits) would be epoch and remaining digit (86 bit) some hash value

input -> string some random string of n size=
output-> 128bit or 38 digits number

39 digits(128 bits)=       epoch +some hashvalue
                    1475166540000+*************************

Could anyone suggest some hash function?

Answer 1

The problem you're having is that 128 bits is not 39 digits, but only 4. If you're willing to change the length of the hash, you can do something like this (total 344 bits):

String input = "abcdef";
String output = null;

try {
    MessageDigest digest = MessageDigest.getInstance("SHA-256");
    byte[] hash = digest.digest(input.getBytes(StandardCharsets.UTF_8));

    output = System.currentTimeMillis() + new String(hash, StandardCharsets.UTF_8);
} catch (NoSuchAlgorithmException e) {
    e.printStackTrace();
}

System.out.println(output);  // Your new hash