blocks|key|3044578|text|您可以重复应用目录名以爬得更高：dirname(dirname(file))。然而，这只能到达根包。如果这是一个问题，请使用os.path.abspath：dirname(dirname(abspath(file)))。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3044579|entityMap^0|G|M|1Q|F|26|V|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

You can apply dirname repeatedly to climb higher: <code>dirname(dirname(file))</code>. This can only go as far as the root package, however. If this is a problem, use <code>os.path.abspath</code>: <code>dirname(dirname(abspath(file)))</code>.

blocks|key|3044665|text|可能是加入两个..文件夹，才能得到父文件夹的父文件夹？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3044666|path+=+os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(__file__)),"..",".."))|code-block|syntax|javascript|3044667|entityMap^0|7|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

May be join two <code>..</code> folder, to get parent of the parent folder?

<pre><code>path = os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(__file__)),"..",".."))
</code></pre>

blocks|key|307910|text|os.path.dirname(os.path.abspath(__file__))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|307911|应该会给你到a的路径。|unstyled|offset|length|style|CODE|307912|但是，如果b.py是当前正在执行的文件，那么您可以通过执行以下命令来实现相同的目的|307913|os.path.abspath(os.path.join('templates',+'blog1',+'page.html'))|307914|entityMap^0|0|6|1|0|5|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@$G|V|H|W|I|J]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|O|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>os.path.dirname(os.path.abspath(__file__))
</code></pre>

Should give you the path to <code>a</code>.

But if <code>b.py</code> is the file that is currently executed, then you can achieve the same by just doing

<pre><code>os.path.abspath(os.path.join('templates', 'blog1', 'page.html'))
</code></pre>

blocks|key|3044707|text|os.path.abspath不验证任何东西，所以如果我们已经将字符串附加到__file__中，那么就不需要费心使用dirname或doesn或其他任何东西了。只需将__file__视为一个目录，并开始攀登：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3044708|#+climb+to+__file__'s+parent's+parent:
os.path.abspath(__file__+%2B+"/../../")|code-block|syntax|javascript|3044709|这远没有os.path.abspath(os.path.join(os.path.dirname(__file__),".."))那么复杂，并且和dirname(dirname(__file__))一样易于管理。超过两层的攀登开始变得荒谬起来。|3044710|但是，因为我们知道要爬多少层，所以我们可以用一个简单的小函数来清理它：|3044711|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'|3044712|entityMap^0|0|F|12|8|1M|7|2B|8|0|0|4|1P|21|Q|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|R|$]]

<code>os.path.abspath</code> doesn't validate anything, so if we're already appending strings to <code>__file__</code> there's no need to bother with <code>dirname</code> or joining or any of that. Just treat <code>__file__</code> as a directory and start climbing:

<pre><code># climb to __file__'s parent's parent:
os.path.abspath(__file__ + "/../../")
</code></pre>

That's far less convoluted than <code>os.path.abspath(os.path.join(os.path.dirname(__file__),".."))</code> and about as manageable as <code>dirname(dirname(__file__))</code>. 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:

<pre><code>uppath = lambda _path, n: os.sep.join(_path.split(os.sep)[:-n])

# __file__ = "/aParent/templates/blog1/page.html"
&gt;&gt;&gt; uppath(__file__, 1)
'/aParent/templates/blog1'
&gt;&gt;&gt; uppath(__file__, 2)
'/aParent/templates'
&gt;&gt;&gt; uppath(__file__, 3)
'/aParent'
</code></pre>

blocks|key|308871|text|一种简单的方法可以是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|308872|import+os
current_dir+=++os.path.abspath(os.path.dirname(__file__))
parent_dir+=+os.path.abspath(current_dir+%2B+"/../")
print+parent_dir|code-block|syntax|javascript|308873|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A simple way can be:

<pre><code>import os
current_dir = os.path.abspath(os.path.dirname(__file__))
parent_dir = os.path.abspath(current_dir + "/../")
print parent_dir
</code></pre>

blocks|key|310960|text|对于../来说，os.pardir是一种更好的方式，而且更具可读性。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|310961|import+os
print+os.path.abspath(os.path.join(given_path,+os.pardir))++|code-block|syntax|javascript|310962|这将返回given_path的父路径|310963|entityMap^0|2|3|8|9|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

<code>os.pardir</code> is a better way for <code>../</code> and more readable. 

<pre><code>import os
print os.path.abspath(os.path.join(given_path, os.pardir)) 
</code></pre>

This will return the parent path of the given_path

blocks|key|310973|text|我认为使用这个更好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|310974|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'|code-block|syntax|javascript|310975|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I think use this is better:

<pre><code>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'
</code></pre>

blocks|key|311020|text|这是另一个相对简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|311021|311022|不使用dirname()+(在一级参数，如"file.txt“或”..“等相对父级参数上不能按预期工作)|unordered-list-item|offset|length|style|CODE|311023|不使用abspath()+(避免任何有关当前工作目录的假设)，而是保留|311024|路径的相对字符|311025|它只使用normpath和join|311026|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)|code-block|syntax|javascript|311027|结果：|311028|.
foo/bar
with/trailing
dir
..
/abs|311029|entityMap^0|0|0|3|9|0|3|9|0|0|4|8|D|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|12|8|@$F|13|G|14|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|15|8|@$F|16|G|17|H|I]]|9|@]|A|$]]|$1|L|3|M|5|6|7|18|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|19|8|@$F|1A|G|1B|H|I]|$F|1C|G|1D|H|I]]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1E|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|R|7|1G|8|@]|9|@]|A|$S|T]]|$1|Y|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|Z|$]]

Here is another relatively simple solution that:

<ul>
<li>does not use <code>dirname()</code> (which does not work as expected on one level arguments like "file.txt" or relative parents like "..")</li>
<li>does not use <code>abspath()</code> (avoiding any assumptions about the current working directory) but instead preserves the relative character of paths</li>
</ul>

it just uses <code>normpath</code> and <code>join</code>:

<pre><code>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)
</code></pre>

Result:

<pre><code>.
foo/bar
with/trailing
dir
..
/abs
</code></pre>

blocks|key|308903|text|我试过了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|308904|import+os
os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))),+os.pardir))|code-block|syntax|javascript|308905|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I tried: 

<pre><code>import os
os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))), os.pardir))
</code></pre>

blocks|key|311043|text|使用以下命令跳转到上一个文件夹：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|311044|os.chdir(os.pardir)|code-block|syntax|javascript|311045|如果您需要多个跳转，在这种情况下使用一个简单的装饰器将是一个好的和简单的解决方案。|311046|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Use the following to jump to previous folder:

<pre><code>os.chdir(os.pardir)
</code></pre>

If you need multiple jumps a good and easy solution will be to use a simple decorator in this case.

blocks|key|3258785|text|将相对路径与Python+3.4%2B中的pathlib模块一起使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|3258786|from+pathlib+import+Path

Path(__file__).parent|code-block|syntax|javascript|3258787|您可以使用多个对parent的调用在path中更进一步：|3258788|Path(__file__).parent.parent|3258789|作为两次指定parent的替代方法，您可以使用：|3258790|Path(__file__).parents[1]|3258791|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/pathlib.html^0|1|4|J|7|J|7|0|0|0|8|6|0|0|6|6|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|D]]|E|@$9|16|A|17|1|18]]|F|$]]|$1|G|3|H|5|I|7|19|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|1A|8|@$9|1B|A|1C|B|D]]|E|@]|F|$]]|$1|N|3|O|5|I|7|1D|8|@]|E|@]|F|$J|K]]|$1|P|3|Q|5|6|7|1E|8|@$9|1F|A|1G|B|D]]|E|@]|F|$]]|$1|R|3|S|5|I|7|1H|8|@]|E|@]|F|$J|K]]|$1|T|3|-4|5|6|7|1I|8|@]|E|@]|F|$]]]|U|$V|$5|W|X|Y|F|$Z|10]]]]

Use relative path with the <a href="https://docs.python.org/3/library/pathlib.html" rel="noreferrer"><code>pathlib</code></a> module in Python 3.4+:

<pre><code>from pathlib import Path

Path(__file__).parent
</code></pre>

You can use multiple calls to <code>parent</code> to go further in the path:

<pre><code>Path(__file__).parent.parent
</code></pre>

As an alternative to specifying <code>parent</code> twice, you can use:

<pre><code>Path(__file__).parents[1]
</code></pre>

this code is get the templates/blog1/page.html in b.py:

<pre><code>path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog1/page.html'))
</code></pre>

but i want to get the parent dir location:

<pre><code>aParent
 |--a
 | |---b.py
 | |---templates
 | |--------blog1
 | |-------page.html
 |--templates
 |--------blog1
 |-------page.html
</code></pre>

and how to get the aParent location

thanks

updated:

this is right:

<pre><code>dirname=os.path.dirname
path = os.path.join(dirname(dirname(__file__)), os.path.join('templates', 'blog1/page.html'))
</code></pre>

or 

<pre><code>path = os.path.abspath(os.path.join(os.path.dirname(__file__),".."))
</code></pre>

How to get the parent dir location

Python

下面的代码是在b.py中获取模板/blog1/page.html：path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog1/page.html'))但是我想要获取父目录的位置：aParent   |--a   |  |...

问如何获取父目录位置
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