需要帮助解压缩存储在Aztec条码中的zlib数据(德国火车票) [英] Need help decompressing zlib data stored in Aztec barcode (Deutsche Bahn Ticket)
问题描述
我正在尝试从Deutsche Bahn(德国铁路)发行的车票上当前使用的条形码格式解码数据.我发现这个非常有用的网站(德语)已经做了类似的事情并提供了 Python脚本.
I'm trying to decode the data from the barcode format currently used on tickets issued by Deutsche Bahn (german railway). I have found this very useful website (german) that already does a similar thing and offers a python script.
网站指出，数据已用zlib压缩，生成的blob已用DSA签名，并且所有数据都存储在条形码(Aztec格式)中.
此类条形码的示例
The website states that the data is compressed with zlib, the resulting blob is signed with DSA and all of it is stored in the barcode (Aztec format).
Example of such a barcode
我已使用网站上提供的脚本成功解码了票证. 安装了python-pyasn1库.读取条形码(按照说明使用 BCTester ，对NeoReader应用有麻烦)并将结果转换为十六进制.将十六进制数据另存为纯文本文件(出于某种原因，由于脚本需要)，并使用脚本解析该文件.奏效了.
I have used the script provided on the website to successfully decode a ticket. Installed the python-pyasn1 library. Read the barcode (used BCTester as per instructions, had some trouble with NeoReader app) and converted the result to hex. Saved the hex data as plain text file (as is for some reason required by the script) and parsed the file with the script. It worked.
但是脚本执行了太多操作.我想自己进行解析，但是无法使zlib解压缩正常工作，并且我几乎不了解其中的代码来理解它. 我几乎不了解Python.不过，我有一些编程经验.
But the script is doing too much. I'd like to do the parsing myself, but I can't get the zlib decompression to work and I understand to little of the code to make sense of it. I know almost no Python. I have some programming experience, though.
如果仅查看条形码中的数据，则看起来像这样: https://gist.github.com/oelna/096787dc18596aaa4f5f
If you simply look at the data from the barcode, it looks like this: https://gist.github.com/oelna/096787dc18596aaa4f5f
第一个问题是: DSA签名是什么，我需要首先将其与实际的压缩数据分开吗?
The first question would be: What is the DSA signature and do I need to split it from the actual compressed data first?
第二个: 一个简单的python脚本看起来像什么，它可以从文件中读取条形码blob并简单地将其解压缩，因此我可以进一步解析该格式.我脑子里有个像
The second: What could a simple python script look like that reads the barcode blob from a file and simply decompresses it, so I can further parse the format. I had something in mind like
#!/usr/bin python
import zlib
ticket = open('ticketdata.txt').read()
print zlib.decompress(ticket)
但是它不起作用.朝正确方向的任何提示将不胜感激.
but it's not working. Any hint in the right direction would be appreciated.
以下是保存到文件中的脚本可以读取的十六进制数据:
Here is the hex data that is readable by the script if saved to a file:
23 55 54 30 31 30 30 38 30 30 30 30 30 31 30 2c 02 14 1c 3d e9 2d cd 5e c4 c0 56 bd ae 61 3e 54 ad a1 b3 26 33 d2 02 14 40 75 03 d0 cf 9c c1 f5 70 58 bd 59 50 a7 af c5 eb 0a f4 74 00 00 00 00 30 32 37 31 78 9c 65 50 cb 4e c3 30 10 e4 53 2c 71 43 4a d9 f5 2b 36 b7 84 04 52 01 55 51 40 1c 51 01 23 2a 42 0e 21 15 3f c7 8d 1f 63 36 11 52 2b 7c f1 78 76 76 66 bd f7 8f 4d 5d 54 c4 44 ce 10 05 d2 eb 78 5b ac 32 7b b4 77 c8 11 6b 62 c7 d6 79 aa ea aa 16 e1 b2 22 4d c4 01 ad 36 58 61 ca 6b 30 c6 e5 64 a0 b6 97 0f a6 a9 6f d6 71 df c7 cf 3e 7f 37 93 66 8e c6 71 de 92 4c c0 e1 22 0d fd 57 7a cb ee b6 cf ef 69 54 fd 66 44 05 31 d0 03 18 01 05 40 04 70 9c 51 46 ad 38 49 33 00 86 20 dd 42 88 04 22 5f a6 a1 db f6 78 79 d4 79 95 76 1f 3f df fd e7 98 86 16 b1 30 0b 65 d6 3c bd 2a 15 ce d8 ab e5 79 9d 47 7b da 34 13 c7 34 73 5a 6b 0b 35 72 d9 5c 0d bb ae 53 aa e8 5f 86 b4 01 e9 25 8d 0d 50 8e 72 3c 39 3c b2 13 94 82 74 ce 2d c7 b3 41 8b ed 4c 9f f5 0b e2 85 6c 01 8c fe c7 b8 e9 87 8c d9 f1 90 28 a3 73 fe 05 6d de 5f f1
更新/解决方案:
马克·阿德勒(Mark Adler)的小技巧使我走上了正确的轨道.这花了我几个小时，但是我为这个特殊问题找到了可行的解决方案.如果我比较聪明，我会在偏移量68处识别zlib头文件78 9C.只需在这一点拆分数据，然后下半部分进行解压缩而不会产生任何抱怨.被警告，非常伤心python
Update/Solution:
Mark Adler's tip set me on the right track. It took me hours, but I hacked together a working solution to this particular problem. If I had been smarter, I would have recognized the zlib header 78 9C at offset 68. Simply split the data at this point and the second half decompresses without complaint. Be warned, very sad python
dsa_signature = ''
zlib_data = ''
cursor = 0
with open('ticketdata.txt', "rb") as fp:
chunk = fp.read(1)
while chunk:
if(cursor < 68):
dsa_signature += chunk
else:
zlib_data += chunk
chunk = fp.read(1)
cursor = cursor + 1
print "\nSignature:"
print "%s\n" % dsa_signature
print "\nCompressed data:"
print "%s\n" % zlib_data
print "\nDecoded:"
print zlib.decompress(zlib_data)
如果有一个简单的解决方案，请随时发表评论.我将继续对此进行更多研究，并尝试使其成为一种更可靠的解决方案，该解决方案可以主动寻找zlib标头，而无需对偏移量进行硬编码.前半部分是一个标识符代码，例如#UT010080000060,，后跟一个ASN.1 DSA签名，幸运的是，我不需要验证或修改.
If there is an easy solution to this, feel free to comment. I'll continue working on this for a little more and try to make it a more robust solution that actively seeks out the zlib header, without hardcoding the offset. The first half is an identifier code, like #UT010080000060,, followed by a ASN.1 DSA signature, which luckily I don't need to verify or modify.
推荐答案
有一个完整有效的zlib流，从十六进制数据中的偏移量68开始，一直到结尾.解压缩为:
There is a complete and valid zlib stream starting at offset 68 in your hex data, and going to the end. It decompresses to:
U_HEAD01005300802P9QAN-4 0501201514560DEDE0080ID0200180104840080080BL020357031204GW3HEMP9 06012015060120151021193517S0010018Fernweh-Ticket natS00200012S0030001AS00900051-0-0S01200010S0140002S2S0150006柏林S0160011NeumünsterS0210038B-Hbf 8:16 ICE794/HH-Hbf 10:16 IC2224S0230013Krull AndreaS026000213S0270019 *************** 0484S0280013Andrea#Krull S031001006.01.2015S032001006.01.2015S035000511160S0360003271
U_HEAD01005300802P9QAN-4������������0501201514560DEDE0080ID0200180104840080BL020357031204GW3HEMP9�����������06012015060120151021193517S0010018Fernweh-Ticket natS00200012S0030001AS00900051-0-0S01200010S0140002S2S0150006BerlinS0160011NeumünsterS0210038B-Hbf 8:16 ICE794/HH-Hbf 10:16 IC2224S0230013Krull AndreaS026000213S0270019***************0484S0280013Andrea#Krull S031001006.01.2015S032001006.01.2015S035000511160S0360003271
如果删除示例的前68个字节，则zlib.decompress()将返回以上内容.
If you drop the first 68 bytes of your example, zlib.decompress() will return the above.
由您决定前68个字节是什么.
It's up to you to figure out what the first 68 bytes are.
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