我想只提取路径的文件名部分.我的代码可以使用,但我想知道更好(pythonic)的方法是什么.
filename = ''
tmppath = '/dir1/dir2/dir3/file.exe'
for i in reversed(tmppath):
if i != '/':
filename += str(i)
else:
break
a = filename[::-1]
print a
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Bar*_*ers 12
尝试:
#!/usr/bin/python
import os.path
path = '/dir1/dir2/dir3/file.exe'
name = os.path.basename(path)
print name
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