blocks|key|927297|text|Python模块(包括Django应用程序)有一个__file__属性，该属性告诉您它们的__init__.py文件在文件系统上的位置，因此|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|927298|import+appname
pth+=+os.path.dirname(appname.__file__)|code-block|syntax|javascript|927299|应该做你想做的事。|927300|在通常情况下，os.path.absname(appname.__path__[0])，但如果应用程序想要以一种奇怪的方式导入文件，则可以更改这一点。|927301|(不过，我总是在我的settings.py中执行PROJECT_ROOT+=+os.path.dirname(os.path.abspath(__file__))+--这使得需要绝对路径的各种设置变得很容易。)|927302|entityMap^0|P|8|19|B|0|0|0|7|10|0|A|B|O|1L|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

Python modules (including Django apps) have a <code>__file__</code> attribute that tells you the location of their <code>__init__.py</code> file on the filesystem, so

<pre><code>import appname
pth = os.path.dirname(appname.__file__)
</code></pre>

should do what you want.

In usual circumstances, <code>os.path.absname(appname.__path__[0])</code>, but it's possible for apps to change that if they want to import files in a weird way.

(I do always do <code>PROJECT_ROOT = os.path.dirname(os.path.abspath(__file__))</code> in my <code>settings.py</code>, though -- makes it easy for the various settings that need to be absolute paths.)

blocks|key|927321|text|因此，公认的答案通常工作得很好。但是，对于|type|unstyled|depth|inlineStyleRanges|entityRanges|data|927322|具有多个路径的|927323|命名空间包，或在配置|unordered-list-item|927324|中显式配置其路径的|927325|应用程序|927326|927327|它们的预期路径可能与模块的__file__属性不一致。|offset|length|style|CODE|927328|Django+(1.7%2B)提供了AppConfig.path属性--我认为即使在简单的情况下，这个属性也更清晰，它也涵盖了这些边缘情况。|927329|The+application+docs告诉您如何获取AppConfig对象。因此，要获取AppConfig并打印其中的路径：|927330|from+django.apps+import+apps
print(apps.get_app_config('app_label').path)|code-block|syntax|javascript|927331|编辑:修改了如何使用get_app_config删除点的示例，这似乎令人困惑。Here+are+the+docs+for+reference。|927332|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/2.0/ref/applications/#django.apps.AppConfig.path|1|https://docs.djangoproject.com/en/2.0/ref/applications/|2|https://docs.djangoproject.com/en/2.2/ref/applications/#django.apps.apps.get_app_config^0|0|0|0|0|0|0|D|8|0|G|E|G|E|0|0|0|K|1|0|0|A|E|13|V|2|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1F|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1G|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1H|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|1I|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1K|8|@$N|1L|O|1M|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|6|7|1N|8|@$N|1O|O|1P|P|Q]]|9|@$N|1Q|O|1R|1|1S]]|A|$]]|$1|T|3|U|5|6|7|1T|8|@]|9|@$N|1U|O|1V|1|1W]]|A|$]]|$1|V|3|W|5|X|7|1X|8|@]|9|@]|A|$Y|Z]]|$1|10|3|11|5|6|7|1Y|8|@$N|1Z|O|20|P|Q]]|9|@$N|21|O|22|1|23]]|A|$]]|$1|12|3|-4|5|6|7|24|8|@]|9|@]|A|$]]]|13|$14|$5|15|16|17|A|$18|19]]|1A|$5|15|16|17|A|$18|1B]]|1C|$5|15|16|17|A|$18|1D]]]]

So the accepted answer usually works fine. However, for

<ul>
<li>namespace packages with multiple paths, or</li>
<li>apps which explicitly configure their paths in the config,</li>
</ul>

their intended path may not agree with the <code>__file__</code> attribute of the module.

Django (1.7+) provides the <a href="https://docs.djangoproject.com/en/2.0/ref/applications/#django.apps.AppConfig.path" rel="noreferrer"><code>AppConfig.path</code></a> attribute - which I think is clearer even in simple cases, and which covers these edge cases too.

<a href="https://docs.djangoproject.com/en/2.0/ref/applications/" rel="noreferrer">The application docs</a> tell you how to get the AppConfig object. So to get AppConfig and print the path from it:

<pre><code>from django.apps import apps
print(apps.get_app_config('app_label').path)
</code></pre>

Edit:
Altered example for how to use <code>get_app_config</code> to remove dots, which seems to have been confusing. <a href="https://docs.djangoproject.com/en/2.2/ref/applications/#django.apps.apps.get_app_config" rel="noreferrer">Here are the docs for reference</a>.

blocks|key|927039|text|通常，这是我在settings.py文件中添加的内容，以便可以引用项目根目录。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|927040|import+os.path

PROJECT_ROOT+=+os.path.abspath(os.path.dirname(__file__))|code-block|syntax|javascript|927041|此方法将获取任何python文件的目录。|927042|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|$]]

Normally, this is what I add in my settings.py file so I can reference the project root.

<pre><code>import os.path

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

This method will get the directory of any python file.

blocks|key|927075|text|在较新版本的Django中(我不知道它是什么时候开始的，我假设它已经存在很多年了)，默认的settingy.py包含一个条目|type|unstyled|depth|inlineStyleRanges|entityRanges|data|927076|#+Build+paths+inside+the+project+like+this:+os.path.join(BASE_DIR,+...)
BASE_DIR+=+os.path.dirname(os.path.dirname(os.path.abspath(__file__)))|code-block|syntax|javascript|927077|您可以通过导入设置文件使用它，如下所示|927078|import+os++++
from+my_project_name.settings+import+BASE_DIR

os.path.join(BASE_DIR,+...)|927079|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

În newer versions of Django (I don't know when it started, I assume it's there for many years now), the default settingy.py contains an entry

<pre><code># Build paths inside the project like this: os.path.join(BASE_DIR, ...)
BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
</code></pre>

which you can use by importing the settings file like this

<pre><code>import os 
from my_project_name.settings import BASE_DIR

os.path.join(BASE_DIR, ...)
</code></pre>

blocks|key|5430210|text|Python3.4及更高版本附带标准库pathlib。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5430211|from+pathlib+import+Path
import+appmodule

pth+=+Path(appmodule.__file__).parent+/+'fixtures'
if+pth.exists():
+++++"Your+code+here"|code-block|syntax|javascript|5430212|parent将为您提供应用程序目录的路径，而/将附加fixtures作为路径。|5430213|entityMap^0|J|7|0|0|0|6|M|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Python3.4 and above comes with standard library <code>pathlib</code>.

<pre><code>from pathlib import Path
import appmodule

pth = Path(appmodule.__file__).parent / 'fixtures'
if pth.exists():
 "Your code here"
</code></pre>

<code>parent</code> will give you the path of your app directory, and <code>/</code> will append fixtures as path.

blocks|key|1389571|text|请记住，appname.__path__是一个列表：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1389572|import+appname
APP_ROOT+=+os.path.abspath(appname.__path__[0])
file_path+=+os.path.join(APP_ROOT,+"some_file.txt")|code-block|syntax|javascript|1389573|entityMap^0|4|G|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|$]]

Keep in mind that <code>appname.__path__</code> is a list:

<pre><code>import appname
APP_ROOT = os.path.abspath(appname.__path__[0])
file_path = os.path.join(APP_ROOT, "some_file.txt")

</code></pre>

I am writing a unit test that needs to access an image file that I put in "fixtures" directory right under my django app directory. I want to open up this image file in my test using relative path, which would require me to get the absolute path of the django app. Is there a way to get the absolute path of the django app?

Get absolute path of django app

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在编写一个单元测试，它需要访问我放在django应用程序目录下的"fixtures“目录中的图像文件。我想在我的测试中使用相对路径打开这个图像文件，这将需要我获得django应用程序的绝对路径。有没有办法获得django应用程序的绝对路径？

问获取django应用的绝对路径
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问获取django应用的绝对路径
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