我有一个字符串,我想用它作为文件名,所以我想用Python删除文件名中不允许的所有字符。
我宁愿严格一点,所以假设我想只保留字母、数字和一小组其他字符,如“_-.()”。”。最优雅的解决方案是什么?
文件名需要在多个操作系统(Windows, Linux和Mac OS)上有效——它是我库中的一个MP3文件,以歌曲标题为文件名,并在3台机器之间共享和备份。
我有一个字符串,我想用它作为文件名,所以我想用Python删除文件名中不允许的所有字符。
我宁愿严格一点,所以假设我想只保留字母、数字和一小组其他字符,如“_-.()”。”。最优雅的解决方案是什么?
文件名需要在多个操作系统(Windows, Linux和Mac OS)上有效——它是我库中的一个MP3文件,以歌曲标题为文件名,并在3台机器之间共享和备份。
当前回答
就像S.Lott回答的那样,你可以看看Django框架如何将字符串转换为有效的文件名。
最新和更新的版本在utils/text.py中,并定义了"get_valid_filename",如下所示:
def get_valid_filename(s):
s = str(s).strip().replace(' ', '_')
return re.sub(r'(?u)[^-\w.]', '', s)
(见https://github.com/django/django/blob/master/django/utils/text.py)
其他回答
This whitelist approach (ie, allowing only the chars present in valid_chars) will work if there aren't limits on the formatting of the files or combination of valid chars that are illegal (like ".."), for example, what you say would allow a filename named " . txt" which I think is not valid on Windows. As this is the most simple approach I'd try to remove whitespace from the valid_chars and prepend a known valid string in case of error, any other approach will have to know about what is allowed where to cope with Windows file naming limitations and thus be a lot more complex.
>>> import string
>>> valid_chars = "-_.() %s%s" % (string.ascii_letters, string.digits)
>>> valid_chars
'-_.() abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
>>> filename = "This Is a (valid) - filename%$&$ .txt"
>>> ''.join(c for c in filename if c in valid_chars)
'This Is a (valid) - filename .txt'
你可以使用re.sub()方法替换任何非“类文件”的东西。但实际上,每个字符都可以是有效的;所以没有预先构建的函数(我相信)来完成它。
import re
str = "File!name?.txt"
f = open(os.path.join("/tmp", re.sub('[^-a-zA-Z0-9_.() ]+', '', str))
将导致/tmp/filename.txt的文件句柄。
这是Windows特定路径的另一个答案,使用简单的替换,没有时髦的模块:
import re
def check_for_illegal_char(input_str):
# remove illegal characters for Windows file names/paths
# (illegal filenames are a superset (41) of the illegal path names (36))
# this is according to windows blacklist obtained with Powershell
# from: https://stackoverflow.com/questions/1976007/what-characters-are-forbidden-in-windows-and-linux-directory-names/44750843#44750843
#
# PS> $enc = [system.Text.Encoding]::UTF8
# PS> $FileNameInvalidChars = [System.IO.Path]::GetInvalidFileNameChars()
# PS> $FileNameInvalidChars | foreach { $enc.GetBytes($_) } | Out-File -FilePath InvalidFileCharCodes.txt
illegal = '\u0022\u003c\u003e\u007c\u0000\u0001\u0002\u0003\u0004\u0005\u0006\u0007\u0008' + \
'\u0009\u000a\u000b\u000c\u000d\u000e\u000f\u0010\u0011\u0012\u0013\u0014\u0015' + \
'\u0016\u0017\u0018\u0019\u001a\u001b\u001c\u001d\u001e\u001f\u003a\u002a\u003f\u005c\u002f'
output_str, _ = re.subn('['+illegal+']','_', input_str)
output_str = output_str.replace('\\','_') # backslash cannot be handled by regex
output_str = output_str.replace('..','_') # double dots are illegal too, or at least a bad idea
output_str = output_str[:-1] if output_str[-1] == '.' else output_str # can't have end of line '.'
if output_str != input_str:
print(f"The name '{input_str}' had invalid characters, "
f"name was modified to '{output_str}'")
return output_str
当测试check_for_illegal_char('fas\u0003\u0004good\\..asd.'),我得到:
The name 'fas♥♦good\..asd.' had invalid characters, name was modified to 'fas__good__asd'
为python 3.6修改的答案
import string
import unicodedata
validFilenameChars = "-_.() %s%s" % (string.ascii_letters, string.digits)
def removeDisallowedFilenameChars(filename):
cleanedFilename = unicodedata.normalize('NFKD', filename).encode('ASCII', 'ignore')
return ''.join(chr(c) for c in cleanedFilename if chr(c) in validFilenameChars)
当遇到同样的问题时,我使用python-slugify。
Shoham也建议使用这种方法,但正如therealmarv指出的那样,默认情况下python-slugify也会转换圆点。
可以通过在regex_pattern参数中包含点来否决这种行为。
> filename = "This is a väryì' Strange File-Nömé.jpeg"
> pattern = re.compile(r'[^-a-zA-Z0-9.]+')
> slugify(filename,regex_pattern=pattern)
'this-is-a-varyi-strange-file-nome.jpeg'
方法复制的正则表达式模式
ALLOWED_CHARS_PATTERN_WITH_UPPERCASE
python-slugify包的slugify.py文件中的全局变量,并扩展为“。”
请记住,像.()这样的特殊字符必须用\转义。
如果您想保留大写字母,请使用小写=False参数。
> filename = "This is a väryì' Strange File-Nömé.jpeg"
> pattern = re.compile(r'[^-a-zA-Z0-9.]+')
> slugify(filename,regex_pattern=pattern, lowercase=False)
'This-is-a-varyi-Strange-File-Nome.jpeg'
这是使用Python 3.8.4和Python -slugify 4.0.1实现的