仍然没有找到一个好的库来生成有效的文件名。注意，在德语、挪威语或法语等语言中，文件名中的特殊字符非常常见，完全可以接受。所以我最终有了自己的图书馆:

# util/files.py

CHAR_MAX_LEN = 31
CHAR_REPLACE = '_'

ILLEGAL_CHARS = [
    '#',  # pound
    '%',  # percent
    '&',  # ampersand
    '{',  # left curly bracket
    '}',  # right curly bracket
    '\\',  # back slash
    '<',  # left angle bracket
    '>',  # right angle bracket
    '*',  # asterisk
    '?',  # question mark
    '/',  # forward slash
    ' ',  # blank spaces
    '$',  # dollar sign
    '!',  # exclamation point
    "'",  # single quotes
    '"',  # double quotes
    ':',  # colon
    '@',  # at sign
    '+',  # plus sign
    '`',  # backtick
    '|',  # pipe
    '=',  # equal sign
]


def generate_filename(
        name, char_replace=CHAR_REPLACE, length=CHAR_MAX_LEN, 
        illegal=ILLEGAL_CHARS, replace_dot=False):
    ''' return clean filename '''
    # init
    _elem = name.split('.')
    extension = _elem[-1].strip()
    _length = length - len(extension) - 1
    label = '.'.join(_elem[:-1]).strip()[:_length]
    filename = ''
    
    # replace '.' ?
    if replace_dot:
        label = label.replace('.', char_replace)
    
    # clean
    for char in label + '.' + extension:
        if char in illegal:
            char = char_replace
        filename += char      
    
    return filename

generate_虚构(“nucgae zutaaer .0.1 docx”，replace_dot=False)

nucgae_zutaäer..0.1.docx

generate_虚构(“nucgae zutaaer .0.1 docx”，replace_dot=True)

nucgae_zutaäer__0_1.docx

为什么不直接用try/except来包装“osopen”，让底层操作系统来判断文件是否有效?

这看起来工作量少得多，而且无论您使用哪种操作系统都是有效的。

请记住，在Unix系统上实际上没有文件名限制

它可能不包含\0 它可能不包含/

其他一切都是公平的。

$ touch "
> even multiline
> haha
> ^[[31m red ^[[0m
> evil"
$ ls -la 
-rw-r--r--       0 Nov 17 23:39 ?even multiline?haha??[31m red ?[0m?evil
$ ls -lab
-rw-r--r--       0 Nov 17 23:39 \neven\ multiline\nhaha\n\033[31m\ red\ \033[0m\nevil
$ perl -e 'for my $i ( glob(q{./*even*}) ){ print $i; } '
./
even multiline
haha
 red 
evil

是的，我只是将ANSI颜色代码存储在一个文件名中，并使它们生效。

为了娱乐，在目录名中放入一个BEL字符，并观看当您CD到其中时所产生的乐趣;)

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)

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。

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'