问题描述
我想根据此处的规范创建一个有效的IFC GUID(IfcGloballyUniqueId): http://www.buildingsmart-tech. org/ifc/IFC2x3/TC1/html/ifcutilityresource/lexical/ifcgloballyuniqueid.htm
I want to create a valid IFC GUID (IfcGloballyUniqueId) according to the specification here:http://www.buildingsmart-tech.org/ifc/IFC2x3/TC1/html/ifcutilityresource/lexical/ifcgloballyuniqueid.htm
它基本上是一个UUID或GUID(128位)，映射到一组22个字符，以限制文本文件中的存储空间.
It's basically a UUID or GUID (128 bit) mapped to a set of 22 characters to limit storage space in a text file.
我目前有此解决方法，但这只是一个近似值:
I currently have this workaround, but it's merely an approximation:
guid = '';22.times{|i|guid<<'0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_$'[rand(64)]}
似乎最好使用ruby SecureRandom生成128位UUID，如本示例中所示("> https://ruby-doc.org/stdlib-2.3.0/libdoc/securerandom/rdoc/SecureRandom.html ):
It seems best to use ruby SecureRandom to generate a 128 bit UUID, like in this example (https://ruby-doc.org/stdlib-2.3.0/libdoc/securerandom/rdoc/SecureRandom.html):
SecureRandom.uuid #=> "2d931510-d99f-494a-8c67-87feb05e1594"
此UUID需要根据以下格式映射到长度为22个字符的字符串:
This UUID needs to be mapped to a string with a length of 22 characters according to this format:
1 2 3 4 5 6
0123456789012345678901234567890123456789012345678901234567890123
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_$";
我不太了解这一点.是否应将32个字符的长十六进制数转换为128个字符的长二进制数，然后划分为22组6位(除了获得剩余2位的一位以外?)，因此可以将每组转换为十进制数从0到64?然后可以用转换表中的相应字符替换哪个字符?
I don't understand this exactly.Should the 32-character long hex-number be converted to a 128-character long binary number, then devided in 22 sets of 6 bits(except for one that gets the remaining 2 bits?) for which each can be converted to a decimal number from 0 to 64? Which then in turn can be replaced by the corresponding character from the conversion table?
我希望有人可以验证我是否在正确的轨道上.
I hope someone can verify if I'm on the right track here.
如果我愿意，那么在Ruby中是否有比使用所有这些单独的转换更快的计算方式来将128位数字转换为22组0-64?
And if I am, is there a computational faster way in Ruby to convert the 128 bit number to the 22 sets of 0-64 than using all these separate conversions?
编辑:对于有相同问题的任何人，这是我目前的解决方案:
Edit: For anyone having the same problem, this is my solution for now:
require 'securerandom'
# possible characters in GUID
guid64 = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_$'
guid = ""
# SecureRandom.uuid: creates a 128 bit UUID hex string
# tr('-', ''): removes the dashes from the hex string
# pack('H*'): converts the hex string to a binary number (high nibble first) (?) is this correct?
# This reverses the number so we end up with the leftover bit on the end, which helps with chopping the sting into pieces.
# It needs to be reversed again to end up with a string in the original order.
# unpack('b*'): converts the binary number to a bit string (128 0's and 1's) and places it into an array
# [0]: gets the first (and only) value from the array
# to_s.scan(/.{1,6}/m): chops the string into pieces 6 characters(bits) with the leftover on the end.
[SecureRandom.uuid.tr('-', '')].pack('H*').unpack('b*')[0].to_s.scan(/.{1,6}/m).each do |num|
# take the number (0 - 63) and find the matching character in guid64, add the found character to the guid string
guid << guid64[num.to_i(2)]
end
guid.reverse
推荐答案
Base64编码与您在此处想要的非常接近，但是映射是不同的.没什么大不了的，您可以解决此问题:
Base64 encoding is pretty close to what you want here, but the mappings are different. No big deal, you can fix that:
require 'securerandom'
require 'base64'
# Define the two mappings here, side-by-side
BASE64 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
IFCB64 = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_$'
def ifcb64(hex)
# Convert from hex to binary, then from binary to Base64
# Trim off the == padding, then convert mappings with `tr`
Base64.encode64([ hex.tr('-', '') ].pack('H*')).gsub(/\=*\n/, '').tr(BASE64, IFCB64)
end
ifcb64(SecureRandom.uuid)
# => "fa9P7E3qJEc1tPxgUuPZHm"
这篇关于如何使用Ruby中的自定义字符集将UUID转换为字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！