[英]converting string to unicode type in python
I'm trying this code: 我正在尝试这段代码:
s = "سلام"
'{:b}'.format(int(s.encode('utf-8').encode('hex'), 16))
but this error occurs: 但是会发生以下错误:
'{:b}'.format(int(s.encode('utf-8').encode('hex'), 16))
UnicodeDecodeError: 'ascii' codec can't decode byte 0xd3 in position 0: ordinal not in range(128)
UnicodeDecodeError:'ascii'编解码器无法解码位置0中的字节0xd3:序数不在范围内(128)
I tried '{:b}'.format(int(s.encode('utf-8').encode('hex'), 16))
but nothing changed. 我试过
'{:b}'.format(int(s.encode('utf-8').encode('hex'), 16))
但没有改变。
what should I do? 我该怎么办?
Since you're using python 2, s = "سلام"
is a byte string (in whatever encoding your terminal uses, presumably utf8): 由于您使用的是python 2,因此
s = "سلام"
是一个字节字符串(无论您的终端使用什么编码,大概是utf8):
>>> s = "سلام"
>>> s
'\xd8\xb3\xd9\x84\xd8\xa7\xd9\x85'
You cannot encode
byte strings (as they are already "encoded"). 您不能
encode
字节字符串(因为它们已经“编码”)。 You're looking for unicode ("real") strings, which in python2 must be prefixed with u
: 你正在寻找unicode(“真实”)字符串,在python2中必须以
u
为前缀:
>>> s = u"سلام"
>>> s
u'\u0633\u0644\u0627\u0645'
>>> '{:b}'.format(int(s.encode('utf-8').encode('hex'), 16))
'1101100010110011110110011000010011011000101001111101100110000101'
If you're getting a byte string from a function such as raw_input
then your string is already encoded - just skip the encode
part: 如果您从诸如
raw_input
的函数获取字节字符串,那么您的字符串已经被编码 - 只需跳过encode
部分:
'{:b}'.format(int(s.encode('hex'), 16))
or (if you're going to do anything else with it) convert it to unicode: 或者(如果你要用它做任何其他事情)将其转换为unicode:
s = s.decode('utf8')
This assumes that your input is UTF-8 encoded, if this might not be the case, check sys.stdin.encoding
first. 这假设您的输入是UTF-8编码,如果情况可能不是这样,请首先检查
sys.stdin.encoding
。
i10n stuff is complicated, here are two articles that will help you further: i10n的内容很复杂,这里有两篇文章可以帮助你进一步:
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