Question

在python中，要创建Type 5 UUID，我们可以简单地执行：

import uuid
print uuid.uuid5(uuid.NAMESPACE_URL, 'my string')

仔细阅读java documentation for java.util.UUID，我不知道该怎么做。首先，没有提到类型5。他们确实有类型3，但签名是：

 nameUUIDFromBytes(byte[] name)
 Static factory to retrieve a type 3 (name based) UUID based on the specified byte array.

我们如何在Java中创建Type 5 UUID？

Answer 1

您可以按照https://stackoverflow.com/a/28776880/1452094中提出的代码自行实施。但是，由于j.u.UUID构造函数需要很长时间，因此需要进行一些调整。

从Java 8开始，标准库似乎不支持类型5.但第三方库喜欢＆＃34; Apache Commons Id＆＃34;有UUID实现支持它。

编辑：这是一个功能齐全的实现：

import java.nio.charset.Charset;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
import java.util.Objects;
import java.util.UUID;

public class UUIDType5 {
    private static final Charset UTF8 = Charset.forName("UTF-8");
    public static final UUID NAMESPACE_DNS = UUID.fromString("6ba7b810-9dad-11d1-80b4-00c04fd430c8");
    public static final UUID NAMESPACE_URL = UUID.fromString("6ba7b811-9dad-11d1-80b4-00c04fd430c8");
    public static final UUID NAMESPACE_OID = UUID.fromString("6ba7b812-9dad-11d1-80b4-00c04fd430c8");
    public static final UUID NAMESPACE_X500 = UUID.fromString("6ba7b814-9dad-11d1-80b4-00c04fd430c8");

    public static UUID nameUUIDFromNamespaceAndString(UUID namespace, String name) {
        return nameUUIDFromNamespaceAndBytes(namespace, Objects.requireNonNull(name, "name == null").getBytes(UTF8));
    }

    public static UUID nameUUIDFromNamespaceAndBytes(UUID namespace, byte[] name) {
        MessageDigest md;
        try {
            md = MessageDigest.getInstance("SHA-1");
        } catch (NoSuchAlgorithmException nsae) {
            throw new InternalError("SHA-1 not supported");
        }
        md.update(toBytes(Objects.requireNonNull(namespace, "namespace is null")));
        md.update(Objects.requireNonNull(name, "name is null"));
        byte[] sha1Bytes = md.digest();
        sha1Bytes[6] &= 0x0f;  /* clear version        */
        sha1Bytes[6] |= 0x50;  /* set to version 5     */
        sha1Bytes[8] &= 0x3f;  /* clear variant        */
        sha1Bytes[8] |= 0x80;  /* set to IETF variant  */
        return fromBytes(sha1Bytes);
    }

    private static UUID fromBytes(byte[] data) {
        // Based on the private UUID(bytes[]) constructor
        long msb = 0;
        long lsb = 0;
        assert data.length >= 16;
        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);
    }

    private static byte[] toBytes(UUID uuid) {
        // inverted logic of fromBytes()
        byte[] out = new byte[16];
        long msb = uuid.getMostSignificantBits();
        long lsb = uuid.getLeastSignificantBits();
        for (int i = 0; i < 8; i++)
            out[i] = (byte) ((msb >> ((7 - i) * 8)) & 0xff);
        for (int i = 8; i < 16; i++)
            out[i] = (byte) ((lsb >> ((15 - i) * 8)) & 0xff);
        return out;
    }
}

要验证它是否有效，我运行了以下代码：

public static void main(String[] args) {
    UUID test = UUIDType5.nameUUIDFromNamespaceAndString(NAMESPACE_URL, "google.com");
    System.out.println(test);
    System.out.println(test.version());
}

这创建了输出：

fedb2fa3-8f5c-5189-80e6-f563dd1cb8f9

5

验证官方python实现：

＆GT;＆GT;＆GT;打印（uuid.uuid5（uuid.NAMESPACE_URL，＆＃39; google.com＆＃39;））

fedb2fa3-8f5c-5189-80e6-f563dd1cb8f9

如何在Java中创建Type 5 UUID？

1 个答案: