由于 glob 不支持正则表达式，您必须蛮力创建匹配字符串.一种方法是利用[]中的字符范围被扩展的事实:

Since glob doesn't support regular expressions, you'll have to brute-force creating the match string. One way is to take advantage of the fact that character ranges in [] are expanded:

C:\temp\py>mkdir 12345678901234

C:\temp\py>C:\Python26\python.exe
Python 2.6.2 Stackless 3.1b3 060516 (release26-maint, Apr 14 2009, 21:19:36) [M
C v.1500 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import glob
>>> glob.glob('./' + ('[0-9]' * 14))
['.\\12345678901234']
>>>

我利用了一个事实，在Python中，将字符串与整数 n 相乘会导致该字符串重复 n 次.

I took advantage of the fact that in Python, multiplying a string with an integer n results in that string being repeated n times.

当然，您可能想继续进行检查，以验证给定的路径实际上是目录:

Of course, you might want to go ahead and put in a check to verify that the given path is actually a directory:

>>> [path for path in glob.iglob('./' + ('[0-9]' * 14))]
['.\\11223344556677', '.\\12345678901234']
>>> [path for path in glob.iglob('./' + ('[0-9]' * 14)) if os.path.isdir(path)]
['.\\12345678901234']