您可以在此处找到格式规范。
A date-time specification.
field octets contents range
----- ------ -------- -----
1 1-2 year* 0..65536
2 3 month 1..12
3 4 day 1..31
4 5 hour 0..23
5 6 minutes 0..59
6 7 seconds 0..60
(use 60 for leap-second)
7 8 deci-seconds 0..9
8 9 direction from UTC '+' / '-'
9 10 hours from UTC* 0..13
10 11 minutes from UTC 0..59
* Notes:
- the value of year is in network-byte order
- daylight saving time in New Zealand is +13 For example,
Tuesday May 26, 1992 at 1:30:15 PM EDT would be displayed as:
1992-5-26,13:30:15.0,-4:0
Note that if only local time is known, then timezone
information (fields 8-10) is not present.
为了解码您的样本数据,您可以使用这个快速而简单的单线:
>>> import struct, datetime
>>> s = '\x07\xd8\t\x17\x03\x184\x00'
>>> datetime.datetime(*struct.unpack('>HBBBBBB', s))
datetime.datetime(2008, 9, 23, 3, 24, 52)
上面的示例远非完美,它没有考虑大小(此对象具有可变大小)并且缺少时区信息。另请注意,字段 7 是分秒 (0..9) 而 timetuple[6] 是微秒 (0 <= x < 1000000);正确的实现留给读者练习。
[更新]
8 年后,让我们尝试解决这个问题(我是懒惰还是什么?):
import struct, pytz, datetime
def decode_snmp_date(octetstr: bytes) -> datetime.datetime:
size = len(octetstr)
if size == 8:
(year, month, day, hour, minutes,
seconds, deci_seconds,
) = struct.unpack('>HBBBBBB', octetstr)
return datetime.datetime(
year, month, day, hour, minutes, seconds,
deci_seconds * 100_000, tzinfo=pytz.utc)
elif size == 11:
(year, month, day, hour, minutes,
seconds, deci_seconds, direction,
hours_from_utc, minutes_from_utc,
) = struct.unpack('>HBBBBBBcBB', octetstr)
offset = datetime.timedelta(
hours=hours_from_utc, minutes=minutes_from_utc)
if direction == b'-':
offset = -offset
return datetime.datetime(
year, month, day, hour, minutes, seconds,
deci_seconds * 100_000, tzinfo=pytz.utc) + offset
raise ValueError("The provided OCTETSTR is not a valid SNMP date")
我不确定我的时区偏移是否正确,但我没有要测试的样本数据,请随时修改答案或在评论中联系我。