获得子目录 [英] Gettings subdirectories
本文介绍了获得子目录的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
您好,
如何获取给定目录的所有子目录? os.listdir()给了我所有条目的
我发现没有办法判断对象是文件还是
目录。
< br $> b $ b谢谢,
Florian
解决方案
Florian Lindner启发我们:< blockquote class =post_quotes>如何获取给定目录的所有子目录?
os.listdir()给我所有条目,我发现没有办法判断对象是否是一个文件或目录。
为什么,你的os.path.isdir()函数不起作用吗?
Sybren
-
世界的问题是愚蠢。并不是说应该对愚蠢的死刑进行处罚,但为什么我们不要仅仅拿掉
安全标签来解决问题呢? br />
Frank Zappa
这有点矫枉过正,但考虑使用os.walk。
root,dirnames,filenames = os.walk(r" C:\")。next()#(在实际代码中,
你想在这里捕获异常)
print dirnames
Florian Lindner写道:您好,
如何获取给定目录的所有子目录? os.listdir()给了我所有条目,我发现没有办法判断对象是文件还是
目录。
谢谢,>
Florian
>> root,dirnames,filenames = os.walk(r" C:\")。next()
哇。这是如何运作的?请指出我可以阅读的内容。我不会在os.walk上看到
。
这很酷。
谢谢,
Rick
Hello,
how can I get all subdirectories of a given directories? os.listdir() gives
me all entries and I''ve found no way to tell if an object is a file or a
directory.
Thanks,
Florian
解决方案
Florian Lindner enlightened us with:how can I get all subdirectories of a given directories?
os.listdir() gives me all entries and I''ve found no way to tell if
an object is a file or a directory.
Why, doesn''t your os.path.isdir() function work?
Sybren
--
The problem with the world is stupidity. Not saying there should be a
capital punishment for stupidity, but why don''t we just take the
safety labels off of everything and let the problem solve itself?
Frank Zappa
It''s a bit overkill, but consider using os.walk.
root, dirnames, filenames = os.walk(r"C:\").next() # (in real code,
you''d want to catch exceptions here)
print dirnames
Florian Lindner wrote:Hello,
how can I get all subdirectories of a given directories? os.listdir() gives
me all entries and I''ve found no way to tell if an object is a file or a
directory.
Thanks,
Florian
>> root, dirnames, filenames = os.walk(r"C:\").next()
Wow. How does that work? Just point me to where I can read about it. I
don''t see it under os.walk.
That''s cool.
Thanks,
Rick
这篇关于获得子目录的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文