如何使用Python对AL32UTF8中的文本进行编码
我们正在尝试使用Python匹配经过Oracle MD5哈希算法处理的哈希。根据他们的说法,在散列之前,所有内容都以AL21UTF8编码:如何使用Python对AL32UTF8中的文本进行编码,python,oracle,Python,Oracle,我们正在尝试使用Python匹配经过Oracle MD5哈希算法处理的哈希。根据他们的说法,在散列之前,所有内容都以AL21UTF8编码: -- Prior to encryption, hashing or keyed hashing, CLOB datatype is -- converted to AL32UTF8. This allows cryptographic data to be -- transferred and understood between databases wi
-- Prior to encryption, hashing or keyed hashing, CLOB datatype is
-- converted to AL32UTF8. This allows cryptographic data to be
-- transferred and understood between databases with different
-- character sets, across character set changes and between
-- separate processes (for example, Java programs).
--
起初我认为UTF-8已经足够好了,但如果我这样做,我的哈希值仍然不匹配。因此,在进一步挖掘之后,我发现了以下内容:
AL32UTF8是适用于XMLType数据的Oracle数据库字符集。它相当于IANA注册的标准UTF-8编码,它支持所有有效的XML字符
请勿将Oracle数据库字符集UTF8(无连字符)与数据库字符集AL32UTF8或字符编码UTF-8混淆。数据库字符集UTF8已被AL32UTF8取代。不要对XML数据使用UTF8。UTF8仅支持Unicode版本3.1及更低版本;它不支持所有有效的XML字符。AL32UTF8没有这样的限制
所以我不能使用UTF-8,也不知道如何让Python的编解码器模块区分UTF-8和utf8。如果我尝试AL32UTF8,它会抛出一个错误。还有人用Python用AL32UTF8编码过吗
我的编解码器代码如下所示:
import codecs
sourceFmt = "ascii"
targetFmt = "utf8"
utfFile = "kesa_utf8.dat"
with codecs.open(old, "rU", sourceFmt) as sourceFile:
with codecs.open(utfFile, "w", targetFmt) as targetFile:
targetFile.write(sourceFile.read())
WC000|IC |KESA |KESA | | | |2012-07-31-15.12.36 |0090| | |\c\n
WC001|100534 |W.47212-0100534 |2012-07-31-15.12.36 | 00000000001270.00|USD|\c\n
WC002|100534 |W.47212-0100534 |Sally |H |Klass |1235 14th St. W. || |Palma Sola ||FL |USA |34209 | | | | | | | | |9412587545 | | |O | | ||20800426|645858741 |SSN | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | || | | | || | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |KESAPC | | | | | |N| | | || | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |\c\n
WC999|1000000000|1000000000|4000000000|
文件本身如下所示:
import codecs
sourceFmt = "ascii"
targetFmt = "utf8"
utfFile = "kesa_utf8.dat"
with codecs.open(old, "rU", sourceFmt) as sourceFile:
with codecs.open(utfFile, "w", targetFmt) as targetFile:
targetFile.write(sourceFile.read())
WC000|IC |KESA |KESA | | | |2012-07-31-15.12.36 |0090| | |\c\n
WC001|100534 |W.47212-0100534 |2012-07-31-15.12.36 | 00000000001270.00|USD|\c\n
WC002|100534 |W.47212-0100534 |Sally |H |Klass |1235 14th St. W. || |Palma Sola ||FL |USA |34209 | | | | | | | | |9412587545 | | |O | | ||20800426|645858741 |SSN | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | || | | | || | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |KESAPC | | | | | |N| | | || | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |\c\n
WC999|1000000000|1000000000|4000000000|
散列应该是86D993FA7121E3B9EE1657A23345FE21
无论如何,我使用hashlib对其进行哈希:
import hashlib
with open(path) as f:
data = f.read()
mdhash = hashlib.md5(data)
mdhash = mdhash.hexdigest()
print mdhash
结果是8421877dd9cdf7235eec47765821998c结果是,无论客户机在做什么,都会导致数据本身发生更改,从而使其具有“\c\n”行结尾,并且在散列后,还会通过填充(末尾的空格)使文件中的行大小相同。一旦我们让客户端停止向我们提供坏数据,我们就能够复制散列。谢谢你的帮助 你能用Oracle中的函数在UTF-8中编码AL32UTF8字符吗?我猜你误解了这篇文章。UTF8有点奇怪,但与您的问题无关。Oracle的AL32UTF8在其他地方被称为UTF-8。请显示您的代码并显示一个示例。ASCII和UTF-8中的相同字符之间没有字节对字节的差异。@Ben-我们自己没有oracle。它来自一个使用oracle的客户机。Codo-I添加了代码和示例