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大家好，

我有稍微修改过的CV1 RSA 证书。所以，我不想使用 asn1wrap 来解析“der”文件，因为它有时会变得过于复杂，而是因为标签已经为 CV1 证书修复，我可以解析HEX 数据这个“der”文件通过将二进制数据转换为十六进制并提取所需的数据范围。

但是对于表示，我希望OID采用点格式， 例如：ABSOLUTE OID 1.3.6.33.4.11.5318.2888.18.10377.5

我可以从整个十六进制数据中提取十六进制字符串： '060D2B0621040BA946964812D10905'

任何可以直接进行此转换的python3函数。或者任何人都可以帮助转换相同的逻辑。

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1 回答 1

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找到任何有兴趣的人的答案。在不使用 pyasn1 或 asn1crypto 的情况下，我没有找到任何将十六进制值转换为 OID 表示法的包。所以我四处浏览，混合了其他语言的代码，并在 python 中创建了一个。

def notation_OID(oidhex_string):

    ''' Input is a hex string and as one byte is 2 charecters i take an 
        empty list and insert 2 characters per element of the list.
        So for a string 'DEADBEEF' it would be ['DE','AD','BE,'EF']. '''
    hex_list = [] 
    for char in range(0,len(oidhex_string),2):
        hex_list.append(oidhex_string[char]+oidhex_string[char+1])

    ''' I have deleted the first two element of the list as my hex string
        includes the standard OID tag '06' and the OID length '0D'. 
        These values are not required for the calculation as i've used 
        absolute OID and not using any ASN.1 modules. Can be removed if you
        have only the data part of the OID in hex string. '''
    del hex_list[0]
    del hex_list[0]

    # An empty string to append the value of the OID in standard notation after
    # processing each element of the list.
    OID_str = ''

    # Convert the list with hex data in str format to int format for 
    # calculations.
    for element in range(len(hex_list)):
        hex_list[element] = int(hex_list[element],16)

    # Convert the OID to its standard notation. Sourced from code in other 
    # languages and adapted for python.

    # The first two digits of the OID are calculated differently from the rest. 
    x = int(hex_list[0] / 40)
    y = int(hex_list[0] % 40)
    if x > 2:
        y += (x-2)*40
        x = 2;

    OID_str += str(x)+'.'+str(y)

    val = 0
    for byte in range(1,len(hex_list)):
        val = ((val<<7) | ((hex_list[byte] & 0x7F)))
        if (hex_list[byte] & 0x80) != 0x80:
            OID_str += "."+str(val)
            val = 0

    # print the OID in dot notation.
    print (OID_str)

notation_OID('060D2B0621040BA946964812D10905')

希望这会有所帮助... cHEERS ！

于 2018-04-04T18:20:23.980 回答