将snmp八位位组字符串转换为人类可读的日期格式 [英] Convert snmp octet string to human readable date format

问题描述

使用pysnmp框架，我在进行snmp漫游时获得了一些值。不幸的是

Using the pysnmp framework i get some values doing a snmp walk. Unfortunately for the oid

1.3.6.1.21.69.1.5.8.1.2（DOCS-CABLE-DEVICE-MIB）

1.3.6.1.21.69.1.5.8.1.2 (DOCS-CABLE-DEVICE-MIB)

我得到一个奇怪的结果，由于它包含ascii字符，如 BEL ，因此我无法在此处正确打印 ACK

i get a weird result which i cant correctly print here since it contains ascii chars like BEL ACK

在执行一次重复时，我得到：

When doing a repr i get:

OctetString（'\x07\xd8\t\x17\x03\x184\x00'）

OctetString('\x07\xd8\t\x17\x03\x184\x00')

但输出应类似于：

2008-9-23,3：24：52.0

2008-9-23,3:24:52.0

该格式称为 DateAndTime。我如何将OctetString输出转换为人类可读的日期/时间？

the format is called "DateAndTime". How can i translate the OctetString output to a "human readable" date/time ?

推荐答案

您可以找到格式说明此处。

You can find the format specification here.

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.

为了解码示例数据，您可以使用以下这种简单又脏的单线： / p>

In order to decode your sample data you can use this quick-and-dirty one-liner:

>>> 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（此对象的大小可变），并且缺少时区信息。还应注意，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。

The example above is far from perfect, it does not account for size (this object has variable size) and is missing timezone information. Also note that the field 7 is deci-seconds (0..9) while timetuple[6] is microseconds (0 <= x < 1000000); the correct implementations is left as an exercise for the reader.

[更新]

8年后，让我们尝试解决此问题（我是懒还是什么？）：

8 years later, lets try to fix this answer (am I lazy or what?):

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")

我不确定时区偏移量是否正确，但我不知道有示例数据可以测试，可以随时修改答案或在评论中查验我。

I'm not sure I got the timezone offset right but I don't have sample data to test, feel free to amend the answer or ping me in the comments.

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