在 Python 中根据用户输入创建保存文件名
我正在编写 IRC 和 XMPP 机器人需要将用户提供的输入转换为文件名。我已经编写了一个函数来执行此操作。够理智吗？
这是代码：
allowednamechars = string.ascii_letters + string.digits + '_+/$.-'
def stripname(name, allowed=""):
""" strip all not allowed chars from name. """
n = name.replace(os.sep, '+')
n = n.replace("@", '+')
n = n.replace("#", '-')
n = n.replace("!", '.')
res = u""
for c in n:
if ord(c) < 31: continue
elif c in allowednamechars + allowed: res += c
else: res += "-" + str(ord(c))
return res
这是一个白名单，其中包含额外的代码，用于删除控制字符并替换 os.sep，以及一些替换以使文件名 Google App Engine 兼容。
相关机器人位于 http://jsonbot.googlecode.com。
那么你觉得怎么样？
I'm programming an IRC and XMPP bot that needs to convert user provided input to a filename. I have already written a function to do this. Is it sane enough?
Here is the code:
allowednamechars = string.ascii_letters + string.digits + '_+/$.-'
def stripname(name, allowed=""):
""" strip all not allowed chars from name. """
n = name.replace(os.sep, '+')
n = n.replace("@", '+')
n = n.replace("#", '-')
n = n.replace("!", '.')
res = u""
for c in n:
if ord(c) < 31: continue
elif c in allowednamechars + allowed: res += c
else: res += "-" + str(ord(c))
return res
It's a whitelist with extra code to remove control characters and replace os.sep, as well as some repaces to make the filename Google App Engine compatible.
The bot in question is at http://jsonbot.googlecode.com.
So what do you think of it?
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urllib.quote(name.encode("utf8"))将产生人类可读的内容，这也应该是安全的。例子：urllib.quote(name.encode("utf8"))will produce something human-readable, which should also be safe. Example:您可能会考虑只执行
base64.urlsafe_b64encode(name)，这将始终生成一个安全的名称，除非您确实想要一个人类可读的文件名。否则，边缘情况的数量会相当长，如果你忘记了其中之一，就会遇到安全问题。You might consider just doing
base64.urlsafe_b64encode(name), which will always produce a safe name, unless you really want a human-readable file name. Otherwise, the number of edge cases is pretty long, and if you forget one of them, you've got a security problem.