问题:如何获取父目录位置
这段代码是在b.py中获取templates / blog1 / page.html:
path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog1/page.html'))
但我想获取父目录位置:
aParent
|--a
| |---b.py
| |---templates
| |--------blog1
| |-------page.html
|--templates
|--------blog1
|-------page.html
以及如何获取父位置
谢谢
更新:
这是对的:
dirname=os.path.dirname
path = os.path.join(dirname(dirname(__file__)), os.path.join('templates', 'blog1/page.html'))
要么
path = os.path.abspath(os.path.join(os.path.dirname(__file__),".."))
this code is get the templates/blog1/page.html in b.py:
path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog1/page.html'))
but i want to get the parent dir location:
aParent
|--a
| |---b.py
| |---templates
| |--------blog1
| |-------page.html
|--templates
|--------blog1
|-------page.html
and how to get the aParent location
thanks
updated:
this is right:
dirname=os.path.dirname
path = os.path.join(dirname(dirname(__file__)), os.path.join('templates', 'blog1/page.html'))
or
path = os.path.abspath(os.path.join(os.path.dirname(__file__),".."))
回答 0
您可以重复应用dirname来爬高:dirname(dirname(file))
。但是,这只能到达根包。如果有问题,请使用os.path.abspath
:dirname(dirname(abspath(file)))
。
You can apply dirname repeatedly to climb higher: dirname(dirname(file))
. This can only go as far as the root package, however. If this is a problem, use os.path.abspath
: dirname(dirname(abspath(file)))
.
回答 1
os.path.abspath
不会验证任何内容,因此,如果我们已经向其追加字符串,__file__
则无需理会dirname
或加入其中的任何一个。只需将其__file__
视为目录并开始爬山:
# climb to __file__'s parent's parent:
os.path.abspath(__file__ + "/../../")
这远不os.path.abspath(os.path.join(os.path.dirname(__file__),".."))
及易处理dirname(dirname(__file__))
。攀登两个以上的水平开始变得荒谬。
但是,由于我们知道要爬多少层,我们可以用一个简单的小函数来清理它:
uppath = lambda _path, n: os.sep.join(_path.split(os.sep)[:-n])
# __file__ = "/aParent/templates/blog1/page.html"
>>> uppath(__file__, 1)
'/aParent/templates/blog1'
>>> uppath(__file__, 2)
'/aParent/templates'
>>> uppath(__file__, 3)
'/aParent'
os.path.abspath
doesn’t validate anything, so if we’re already appending strings to __file__
there’s no need to bother with dirname
or joining or any of that. Just treat __file__
as a directory and start climbing:
# climb to __file__'s parent's parent:
os.path.abspath(__file__ + "/../../")
That’s far less convoluted than os.path.abspath(os.path.join(os.path.dirname(__file__),".."))
and about as manageable as dirname(dirname(__file__))
. Climbing more than two levels starts to get ridiculous.
But, since we know how many levels to climb, we could clean this up with a simple little function:
uppath = lambda _path, n: os.sep.join(_path.split(os.sep)[:-n])
# __file__ = "/aParent/templates/blog1/page.html"
>>> uppath(__file__, 1)
'/aParent/templates/blog1'
>>> uppath(__file__, 2)
'/aParent/templates'
>>> uppath(__file__, 3)
'/aParent'
回答 2
将相对路径与pathlib
Python 3.4+中的模块一起使用:
from pathlib import Path
Path(__file__).parent
您可以使用多个调用来parent
进一步进行以下操作:
Path(__file__).parent.parent
作为指定parent
两次的替代方法,可以使用:
Path(__file__).parents[1]
Use relative path with the pathlib
module in Python 3.4+:
from pathlib import Path
Path(__file__).parent
You can use multiple calls to parent
to go further in the path:
Path(__file__).parent.parent
As an alternative to specifying parent
twice, you can use:
Path(__file__).parents[1]
回答 3
os.path.dirname(os.path.abspath(__file__))
应该给你通往的道路a
。
但是,如果b.py
当前执行的是文件,则只需执行以下操作即可
os.path.abspath(os.path.join('templates', 'blog1', 'page.html'))
os.path.dirname(os.path.abspath(__file__))
Should give you the path to a
.
But if b.py
is the file that is currently executed, then you can achieve the same by just doing
os.path.abspath(os.path.join('templates', 'blog1', 'page.html'))
回答 4
os.pardir
是一种更好的方法../
,更具可读性。
import os
print os.path.abspath(os.path.join(given_path, os.pardir))
这将返回给定路径的父路径
os.pardir
is a better way for ../
and more readable.
import os
print os.path.abspath(os.path.join(given_path, os.pardir))
This will return the parent path of the given_path
回答 5
一种简单的方法可以是:
import os
current_dir = os.path.abspath(os.path.dirname(__file__))
parent_dir = os.path.abspath(current_dir + "/../")
print parent_dir
A simple way can be:
import os
current_dir = os.path.abspath(os.path.dirname(__file__))
parent_dir = os.path.abspath(current_dir + "/../")
print parent_dir
回答 6
可能是加入两个..
文件夹,以获取父文件夹的父文件夹?
path = os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(__file__)),"..",".."))
May be join two ..
folder, to get parent of the parent folder?
path = os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(__file__)),"..",".."))
回答 7
使用以下命令跳到上一个文件夹:
os.chdir(os.pardir)
如果您需要多次跳转,那么在这种情况下,使用简单的装饰器是一个好而简单的解决方案。
Use the following to jump to previous folder:
os.chdir(os.pardir)
If you need multiple jumps a good and easy solution will be to use a simple decorator in this case.
回答 8
这是另一个相对简单的解决方案:
- 不使用
dirname()
(在“ file.txt”这样的一级参数或“ ..”这样的相对父级上不能按预期工作) - 不使用
abspath()
(避免对当前工作目录进行任何假设),而是保留路径的相对字符
它只是使用normpath
和join
:
def parent(p):
return os.path.normpath(os.path.join(p, os.path.pardir))
# Example:
for p in ['foo', 'foo/bar/baz', 'with/trailing/slash/',
'dir/file.txt', '../up/', '/abs/path']:
print parent(p)
结果:
.
foo/bar
with/trailing
dir
..
/abs
Here is another relatively simple solution that:
- does not use
dirname()
(which does not work as expected on one level arguments like “file.txt” or relative parents like “..”) - does not use
abspath()
(avoiding any assumptions about the current working directory) but instead preserves the relative character of paths
it just uses normpath
and join
:
def parent(p):
return os.path.normpath(os.path.join(p, os.path.pardir))
# Example:
for p in ['foo', 'foo/bar/baz', 'with/trailing/slash/',
'dir/file.txt', '../up/', '/abs/path']:
print parent(p)
Result:
.
foo/bar
with/trailing
dir
..
/abs
回答 9
我认为用这个更好:
os.path.realpath(__file__).rsplit('/', X)[0]
In [1]: __file__ = "/aParent/templates/blog1/page.html"
In [2]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[3]: '/aParent'
In [4]: __file__ = "/aParent/templates/blog1/page.html"
In [5]: os.path.realpath(__file__).rsplit('/', 1)[0]
Out[6]: '/aParent/templates/blog1'
In [7]: os.path.realpath(__file__).rsplit('/', 2)[0]
Out[8]: '/aParent/templates'
In [9]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[10]: '/aParent'
I think use this is better:
os.path.realpath(__file__).rsplit('/', X)[0]
In [1]: __file__ = "/aParent/templates/blog1/page.html"
In [2]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[3]: '/aParent'
In [4]: __file__ = "/aParent/templates/blog1/page.html"
In [5]: os.path.realpath(__file__).rsplit('/', 1)[0]
Out[6]: '/aParent/templates/blog1'
In [7]: os.path.realpath(__file__).rsplit('/', 2)[0]
Out[8]: '/aParent/templates'
In [9]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[10]: '/aParent'
回答 10
我试过了:
import os
os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))), os.pardir))
I tried:
import os
os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))), os.pardir))
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