你可以看看Django框架(但要考虑到许可!)，看看他们如何从任意文本中创建一个“slug”。段代码是URL和文件名友好的。

Django文本utils定义了一个函数，slugify()，这可能是这种事情的黄金标准。本质上，他们的代码如下。

import unicodedata
import re

def slugify(value, allow_unicode=False):
    """
    Taken from https://github.com/django/django/blob/master/django/utils/text.py
    Convert to ASCII if 'allow_unicode' is False. Convert spaces or repeated
    dashes to single dashes. Remove characters that aren't alphanumerics,
    underscores, or hyphens. Convert to lowercase. Also strip leading and
    trailing whitespace, dashes, and underscores.
    """
    value = str(value)
    if allow_unicode:
        value = unicodedata.normalize('NFKC', value)
    else:
        value = unicodedata.normalize('NFKD', value).encode('ascii', 'ignore').decode('ascii')
    value = re.sub(r'[^\w\s-]', '', value.lower())
    return re.sub(r'[-\s]+', '-', value).strip('-_')

旧版本是:

def slugify(value):
    """
    Normalizes string, converts to lowercase, removes non-alpha characters,
    and converts spaces to hyphens.
    """
    import unicodedata
    value = unicodedata.normalize('NFKD', value).encode('ascii', 'ignore')
    value = unicode(re.sub('[^\w\s-]', '', value).strip().lower())
    value = unicode(re.sub('[-\s]+', '-', value))
    # ...
    return value

还有更多，但我把它省略了，因为它没有解决怠惰，而是逃避。

给，这应该涵盖了所有的基础。它为您处理所有类型的问题，包括(但不限于)字符替换。

适用于Windows、*nix和几乎所有其他文件系统。只允许打印字符。

def txt2filename(txt, chr_set='normal'):
    """Converts txt to a valid Windows/*nix filename with printable characters only.

    args:
        txt: The str to convert.
        chr_set: 'normal', 'universal', or 'inclusive'.
            'universal':    ' -.0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
            'normal':       Every printable character exept those disallowed on Windows/*nix.
            'extended':     All 'normal' characters plus the extended character ASCII codes 128-255
    """

    FILLER = '-'

    # Step 1: Remove excluded characters.
    if chr_set == 'universal':
        # Lookups in a set are O(n) vs O(n * x) for a str.
        printables = set(' -.0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz')
    else:
        if chr_set == 'normal':
            max_chr = 127
        elif chr_set == 'extended':
            max_chr = 256
        else:
            raise ValueError(f'The chr_set argument may be normal, extended or universal; not {chr_set=}')
        EXCLUDED_CHRS = set(r'<>:"/\|?*')               # Illegal characters in Windows filenames.
        EXCLUDED_CHRS.update(chr(127))                  # DEL (non-printable).
        printables = set(chr(x)
                         for x in range(32, max_chr)
                         if chr(x) not in EXCLUDED_CHRS)
    result = ''.join(x if x in printables else FILLER   # Allow printable characters only.
                     for x in txt)

    # Step 2: Device names, '.', and '..' are invalid filenames in Windows.
    DEVICE_NAMES = 'CON,PRN,AUX,NUL,COM1,COM2,COM3,COM4,' \
                   'COM5,COM6,COM7,COM8,COM9,LPT1,LPT2,' \
                   'LPT3,LPT4,LPT5,LPT6,LPT7,LPT8,LPT9,' \
                   'CONIN$,CONOUT$,..,.'.split()        # This list is an O(n) operation.
    if result in DEVICE_NAMES:
        result = f'-{result}-'

    # Step 3: Maximum length of filename is 255 bytes in Windows and Linux (other *nix flavors may allow longer names).
    result = result[:255]

    # Step 4: Windows does not allow filenames to end with '.' or ' ' or begin with ' '.
    result = re.sub(r'^[. ]', FILLER, result)
    result = re.sub(r' $', FILLER, result)

    return result

这个解决方案不需要外部库。它也替代了不可打印的文件名，因为它们并不总是容易处理。

Github上有个不错的项目叫python-slugify:

安装:

pip install python-slugify

然后使用:

>>> from slugify import slugify
>>> txt = "This\ is/ a%#$ test ---"
>>> slugify(txt)
'this-is-a-test'

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'

我相信这不是一个很好的答案，因为它修改了它循环的字符串，但它似乎工作得很好:

import string
for chr in your_string:
 if chr == ' ':
   your_string = your_string.replace(' ', '_')
 elif chr not in string.ascii_letters or chr not in string.digits:
    your_string = your_string.replace(chr, '')

其他注释还没有解决的另一个问题是空字符串，这显然不是一个有效的文件名。你也可能因为剥离太多字符而得到一个空字符串。

对于Windows保留文件名和点的问题，最安全的回答是“如何从任意用户输入规范化有效文件名?”就是“连试都不用试”:如果你能找到任何其他方法来避免它(例如。使用来自数据库的整数主键作为文件名)，这样做。

如果你必须，你真的需要允许空格和。’将文件扩展名作为名称的一部分，可以这样尝试:

import re
badchars= re.compile(r'[^A-Za-z0-9_. ]+|^\.|\.$|^ | $|^$')
badnames= re.compile(r'(aux|com[1-9]|con|lpt[1-9]|prn)(\.|$)')

def makeName(s):
    name= badchars.sub('_', s)
    if badnames.match(name):
        name= '_'+name
    return name

即使这样也不能保证正确，特别是在意想不到的操作系统上-例如RISC操作系统讨厌空格并使用'。'作为目录分隔符。