Python 2.7，将utf8字符串转换为ascii [英] Python 2.7, convert utf8 string to ascii

问题描述

我正在使用python 2.7.12 我有一个包含Unicode文字的字符串，该文字不是Unicode类型的.我想将其转换为文本.这个例子说明了我要做什么.

I am working with python 2.7.12 I have string which contains a unicode literal, which is not of type Unicode. I would like to convert this to text. This example explains what I am trying to do.

>>> s
'\x00u\x00s\x00e\x00r\x00n\x00a\x00m\x00e\x00'
>>> print s
username
>>> type(s)
<type 'str'>
>>> s == "username"
False

我将如何转换此字符串?

How would I go about converting this string?

推荐答案

这不是UTF-8，而是UTF-16，尽管目前尚不清楚它是大字节序还是小字节序(您没有BOM，并且您有一个前导和尾随的NUL字节，使其长度不均匀).对于ASCII范围内的文本，UTF-8与ASCII是无法区分的，而UTF-16将NUL字节与ASCII编码的字节交替(如您的示例).

That's not UTF-8, it's UTF-16, though it's unclear whether it's big endian or little endian (you have no BOM, and you have a leading and trailing NUL byte, making it an uneven length). For text in the ASCII range, UTF-8 is indistinguishable from ASCII, while UTF-16 alternates NUL bytes with the ASCII encoded bytes (as in your example).

无论如何，转换为纯ASCII是相当容易的，您只需要以一种或另一种方式处理不均匀的长度:

In any event, converting to plain ASCII is fairly easy, you just need to deal with the uneven length one way or another:

s = 'u\x00s\x00e\x00r\x00n\x00a\x00m\x00e\x00' # I removed \x00 from beginning manually
sascii = s.decode('utf-16-le').encode('ascii')

# Or without manually removing leading \x00
sascii = s.decode('utf-16-be', errors='ignore').encode('ascii')

当然，如果您的输入只是NUL散布的ASCII，并且您无法弄清楚字节顺序或如何获得偶数个字节，则可以作弊:

Course, if your inputs are just NUL interspersed ASCII and you can't figure out the endianness or how to get an even number of bytes, you can just cheat:

sascii = s.replace('\x00', '')

但是，如果输入是完全不同的编码，这不会引发异常，因此它可能会隐藏指定您期望捕获的错误.

But that won't raise exceptions in the case where the input is some completely different encoding, so it may hide errors that specifying what you expect would have caught.

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