这个Java UUID5实现没有通过unit testing

我找不到Java的自包含UUID5实现，所以我尝试在下面推广这个解决方案。 它通过我的一些unit testing，但其他人失败。 这有什么明显的错误吗？

public static UUID UUID5(UUID namespace, String name) { MessageDigest md; try { md = MessageDigest.getInstance("SHA-1"); } catch (NoSuchAlgorithmException nsae) { throw new InternalError("SHA-1 not supported"); } byte[] namespaceBytes = ByteBuffer.allocate(16).putLong(namespace.getMostSignificantBits()).putLong(namespace.getLeastSignificantBits()).array(); byte[] nameBytes; try { nameBytes = name.getBytes("UTF-8"); } catch (UnsupportedEncodingException e) { throw new InternalError("UTF-8 encoding not supported"); } byte[] toHashify = new byte[namespaceBytes.length + nameBytes.length]; System.arraycopy(namespaceBytes, 0, toHashify, 0, namespaceBytes.length); System.arraycopy(nameBytes, 0, toHashify, namespaceBytes.length, nameBytes.length); byte[] data = md.digest(toHashify); data = Arrays.copyOfRange(data, 0, 16); // Wikipedia says "Note that the 160 bit SHA-1 hash is truncated to 128 bits to make the length work out." data[6] &= 0x0f; /* clear version */ data[6] |= 0x50; /* set to version 5 TODO is this the correct way to do it */ data[8] &= 0x3f; /* clear variant */ data[8] |= 0x80; /* set to IETF variant */ long msb = 0; long lsb = 0; assert data.length == 16 : "data must be 16 bytes in length"; for (int i=0; i<8; i++) {msb = (msb << 8) | (data[i] & 0xff);} for (int i=8; i<16; i++) {lsb = (lsb << 8) | (data[i] & 0xff);} return new UUID(msb, lsb); } 

Answer 1

使用Apache Commons Id ，更具体地说是UUID.nameUUIDFromString 。

 UUID uuid5 = UUID.nameUUIDFromString("www.ford.com", UUID.fromString("078d4e79-244f-440e-844d-9454eadfd411"), UUID.SHA1_ENCODING); 

UUID.SHA1_ENCODING将生成版本5 UUID。