python - 来自 Python 中 UTC 时间戳的 UUID1？

Question

问题是这样的:

我的应用程序部署在具有不同时区的远程服务器上，我想针对 UTC 时间戳生成一个 uuid1。我找不到从任何给定时间戳生成 uuid1 的方法。我想这样做的原因是我不想陷入计算本地时间的麻烦，因为我的本地时间不遵守夏令时，而远程服务器遵守夏令时，结果表示逻辑变得很麻烦。

限制是Timestamp需要存储为uuid1。对此的任何想法或解决方法将不胜感激。

Answer 1

如果您给它正确的片段，UUID 类将进行位杂耍 - http://docs.python.org/library/uuid.html

要获得正确的组件，您可以从 python2.7 复制 uuid1 代码:

def uuid1(node=None, clock_seq=None):
        """Generate a UUID from a host ID, sequence number, and the current time.
        If 'node' is not given, getnode() is used to obtain the hardware
        address.  If 'clock_seq' is given, it is used as the sequence number;
        otherwise a random 14-bit sequence number is chosen."""
    # When the system provides a version-1 UUID generator, use it (but don't
    # use UuidCreate here because its UUIDs don't conform to RFC 4122).
    if _uuid_generate_time and node is clock_seq is None:
        _buffer = ctypes.create_string_buffer(16)
        _uuid_generate_time(_buffer)
        return UUID(bytes=_buffer.raw)
    global _last_timestamp
    import time
    nanoseconds = int(time.time() * 1e9)
    # 0x01b21dd213814000 is the number of 100-ns intervals between the
    # UUID epoch 1582-10-15 00:00:00 and the Unix epoch 1970-01-01 00:00:00.
    timestamp = int(nanoseconds//100) + 0x01b21dd213814000L
    if _last_timestamp is not None and timestamp <= _last_timestamp:
        timestamp = _last_timestamp + 1
    _last_timestamp = timestamp
    if clock_seq is None:
        import random
        clock_seq = random.randrange(1<<14L) # instead of stable storage
    time_low = timestamp & 0xffffffffL
    time_mid = (timestamp >> 32L) & 0xffffL
    time_hi_version = (timestamp >> 48L) & 0x0fffL
    clock_seq_low = clock_seq & 0xffL
    clock_seq_hi_variant = (clock_seq >> 8L) & 0x3fL
    if node is None:
        node = getnode()
    return UUID(fields=(time_low, time_mid, time_hi_version,
                        clock_seq_hi_variant, clock_seq_low, node), version=1)

所有你需要做的就是复制+粘贴并修改时间戳部分以使用固定值(如果你知道你的时间是不同的，你可以忽略 last_timestamp 部分 - 这只是为了避免时钟分辨率为不足)。