blocks|key|310908|text|您不能直接确定正在执行的主脚本的位置。毕竟，有时候脚本根本不是来自一个文件。例如，它可能来自交互式解释器或仅存储在内存中的动态生成的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|310909|但是，您可以可靠地确定模块的位置，因为模块始终是从文件加载的。如果您使用以下代码创建一个模块，并将其放在与您的主脚本相同的目录中，那么主脚本可以导入该模块并使用它来定位自己。|310910|某些_path/module_locator.py：|310911|def+we_are_frozen():
++++#+All+of+the+modules+are+built-in+to+the+interpreter,+e.g.,+by+py2exe
++++return+hasattr(sys,+"frozen")

def+module_path():
++++encoding+=+sys.getfilesystemencoding()
++++if+we_are_frozen():
++++++++return+os.path.dirname(unicode(sys.executable,+encoding))
++++return+os.path.dirname(unicode(__file__,+encoding))|code-block|syntax|javascript|310912|某些路径/main.py：|310913|import+module_locator
my_path+=+module_locator.module_path()|310914|如果在不同的目录中有几个主脚本，则可能需要多个module_locator副本。|310915|当然，如果您的主脚本是由其他工具加载的，而这些工具不允许您导入与您的脚本位于同一位置的模块，那么您就不走运了。在这种情况下，您要查找的信息根本不存在于程序中的任何地方。最好的办法是向该工具的作者提交一个bug。|310916|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

You can't directly determine the location of the main script being executed. After all, sometimes the script didn't come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.

However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.

some_path/module_locator.py:

<pre><code>def we_are_frozen():
 # All of the modules are built-in to the interpreter, e.g., by py2exe
 return hasattr(sys, "frozen")

def module_path():
 encoding = sys.getfilesystemencoding()
 if we_are_frozen():
 return os.path.dirname(unicode(sys.executable, encoding))
 return os.path.dirname(unicode(__file__, encoding))
</code></pre>

some_path/main.py:

<pre><code>import module_locator
my_path = module_locator.module_path()
</code></pre>

If you have several main scripts in different directories, you may need more than one copy of module_locator.

Of course, if your main script is loaded by some other tool that doesn't let you import modules that are co-located with your script, then you're out of luck. In cases like that, the information you're after simply doesn't exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.

blocks|key|308703|text|简而言之，没有保证获得你想要的信息的方法，但是有一些启发式方法在实践中几乎总是有效的。你可以看看。它从C的角度讨论了这个问题，但是所提出的解决方案很容易被转录成Python。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|308704|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The short answer is that there is no guaranteed way to get the information you want, however there are heuristics that work almost always in practice. You might look at <a href="https://stackoverflow.com/questions/933850/how-to-find-the-location-of-the-executable-in-c">How do I find the location of the executable in C?</a>. It discusses the problem from a C point of view, but the proposed solutions are easily transcribed into Python.

blocks|key|313855|text|我遇到了一个类似的问题，我认为这可能会解决问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|313856|def+module_path(local_function):
+++'''+returns+the+module+path+without+the+use+of+__file__.++Requires+a+function+defined
+++locally+in+the+module.
+++from+http://stackoverflow.com/questions/729583/getting-file-path-of-imported-module'''
+++return+os.path.abspath(inspect.getsourcefile(local_function))|code-block|syntax|javascript|313857|它适用于常规脚本和IDLE。我所能说的就是为其他人尝试一下！|offset|length|313858|我的典型用法是：|313859|from+toolbox+import+module_path
def+main():
+++pass+#+Do+stuff

global+__modpath__
__modpath__+=+module_path(main)|313860|现在我使用__modpath__而不是__file__。|313861|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/IDLE^0|0|0|9|4|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@$I|11|J|12|1|13]]|A|$]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

I was running into a similar problem, and I think this might solve the problem:
<pre><code>def module_path(local_function):
 ''' returns the module path without the use of __file__. Requires a function defined
 locally in the module.
 from http://stackoverflow.com/questions/729583/getting-file-path-of-imported-module'''
 return os.path.abspath(inspect.getsourcefile(local_function))
</code></pre>
It works for regular scripts and in <a href="https://en.wikipedia.org/wiki/IDLE" rel="nofollow noreferrer">IDLE</a>. All I can say is try it out for others!
My typical usage:
<pre><code>from toolbox import module_path
def main():
 pass # Do stuff

global __modpath__
__modpath__ = module_path(main)
</code></pre>
Now I use _modpath_ instead of _file_.

blocks|key|3050505|text|首先，您需要从inspect和os导入|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3050506|from+inspect+import+getsourcefile
from+os.path+import+abspath|code-block|syntax|javascript|3050507|接下来，无论您想从哪里找到源文件，只需使用|3050508|abspath(getsourcefile(lambda:0))|3050509|entityMap^0|7|7|F|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

First, you need to import from <code>inspect</code> and <code>os</code>

<pre><code>from inspect import getsourcefile
from os.path import abspath
</code></pre>

Next, wherever you want to find the source file from you just use

<pre><code>abspath(getsourcefile(lambda:0))
</code></pre>

blocks|key|3050517|text|这应该会以一种跨平台的方式做到这一点(只要你没有使用解释器之类的东西)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3050518|import+os,+sys
non_symbolic=os.path.realpath(sys.argv[0])
program_filepath=os.path.join(sys.path[0],+os.path.basename(non_symbolic))|code-block|syntax|javascript|3050519|sys.path[0]是您的调用脚本所在的目录(它首先在这里查找该脚本要使用的模块)。我们可以将文件本身的名称从sys.argv[0]的末尾去掉(这就是我对os.path.basename所做的)。os.path.join只是以一种跨平台的方式将它们结合在一起。os.path.realpath只是确保如果我们获得的任何符号链接的名称与脚本本身不同，我们仍然可以获得脚本的真实名称。|offset|length|style|CODE|3050520|我没有Mac；所以，我还没有在Mac上测试过。请让我知道，如果它的工作，因为它似乎应该。我在使用Python3.4的Linux+(Xubuntu)中对此进行了测试。请注意，这个问题的许多解决方案在Mac上都不起作用(因为我听说Mac上没有__file__+)。|3050521|请注意，如果您的脚本是一个符号链接，它将为您提供它所链接的文件的路径(而不是符号链接的路径)。|3050522|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Xubuntu^0|0|0|0|B|1K|B|26|G|2R|C|3N|G|0|3B|8|1T|7|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]|$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1B|8|@$I|1C|J|1D|K|L]]|9|@$I|1E|J|1F|1|1G]]|A|$]]|$1|O|3|P|5|6|7|1H|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

This should do the trick in a cross-platform way (so long as you're not using the interpreter or something):
<pre><code>import os, sys
non_symbolic=os.path.realpath(sys.argv[0])
program_filepath=os.path.join(sys.path[0], os.path.basename(non_symbolic))
</code></pre>
<code>sys.path[0]</code> is the directory that your calling script is in (the first place it looks for modules to be used by that script). We can take the name of the file itself off the end of <code>sys.argv[0]</code> (which is what I did with <code>os.path.basename</code>). <code>os.path.join</code> just sticks them together in a cross-platform way. <code>os.path.realpath</code> just makes sure if we get any symbolic links with different names than the script itself that we still get the real name of the script.
I don't have a Mac; so, I haven't tested this on one. Please let me know if it works, as it seems it should. I tested this in Linux (<a href="https://en.wikipedia.org/wiki/Xubuntu" rel="nofollow noreferrer">Xubuntu</a>) with Python 3.4. Note that many solutions for this problem don't work on Macs (since I've heard that <code>__file__</code> is not present on Macs).
Note that if your script is a symbolic link, it will give you the path of the file it links to (and not the path of the symbolic link).

blocks|key|313910|text|有关相关信息，请参阅我对问题Importing+modules+from+parent+folder的回答，包括为什么我的回答没有使用不可靠的__file__变量。这个简单的解决方案应该与不同的操作系统交叉兼容，因为os和inspect模块是Python的一部分。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|313911|首先，您需要导入inspect和os模块的一部分。|BOLD|313912|from+inspect+import+getsourcefile
from+os.path+import+abspath|code-block|syntax|javascript|313913|接下来，在Python代码中需要的其他地方使用下面这一行：|313914|abspath(getsourcefile(lambda:0))|313915|它的工作原理：|313916|从内置模块os+(如下所述)导入abspath工具。|313917|适用于Mac、NT或Posix的|313918|+OS例程取决于我们所在的系统。|blockquote|313919|313920|313921|然后从内置模块inspect导入getsourcefile+(如下所述)。|313922|313923|从活动的Python对象中获取有用的信息。|313924|313925|313926|313927|abspath(path)返回文件的绝对/完整版本path|unordered-list-item|313928|getsourcefile(lambda:0)以某种方式获取lambda函数对象的内部源文件，因此返回Python+shell中的'<pyshell#nn>'或返回当前正在执行的Python代码的文件路径。|313929|313930|对getsourcefile(lambda:0)的结果使用abspath应该确保生成的文件路径是Python文件的完整文件路径。|313931|这个解释过的解决方案最初是基于How+do+I+get+the+path+of+the+current+executed+file+in+Python?的答案中的代码。|313932|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/714063/importing-modules-from-parent-folder/33532002#33532002|1|https://stackoverflow.com/questions/2632199/how-do-i-get-the-path-of-the-current-executed-file-in-python/18489147#18489147^0|1Z|8|30|2|33|7|E|10|0|0|8|7|G|2|0|0|0|0|0|5|2|G|7|0|0|0|0|0|7|7|G|D|0|0|0|0|0|0|0|D|0|0|N|1T|E|0|0|1|N|T|7|0|F|1P|1|0^^$0|@$1|2|3|4|5|6|7|1U|8|@$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]]|D|@$9|21|A|22|1|23]]|E|$]]|$1|F|3|G|5|6|7|24|8|@$9|25|A|26|B|H]|$9|27|A|28|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|29|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|2A|8|@]|D|@]|E|$]]|$1|P|3|Q|5|K|7|2B|8|@]|D|@]|E|$L|M]]|$1|R|3|S|5|6|7|2C|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|2D|8|@$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|2I|8|@]|D|@]|E|$]]|$1|X|3|Y|5|Z|7|2J|8|@]|D|@]|E|$]]|$1|10|3|-4|5|6|7|2K|8|@]|D|@]|E|$]]|$1|11|3|-4|5|6|7|2L|8|@]|D|@]|E|$]]|$1|12|3|13|5|6|7|2M|8|@$9|2N|A|2O|B|C]|$9|2P|A|2Q|B|C]]|D|@]|E|$]]|$1|14|3|-4|5|6|7|2R|8|@]|D|@]|E|$]]|$1|15|3|16|5|Z|7|2S|8|@]|D|@]|E|$]]|$1|17|3|-4|5|6|7|2T|8|@]|D|@]|E|$]]|$1|18|3|-4|5|6|7|2U|8|@]|D|@]|E|$]]|$1|19|3|-4|5|6|7|2V|8|@]|D|@]|E|$]]|$1|1A|3|1B|5|1C|7|2W|8|@$9|2X|A|2Y|B|C]]|D|@]|E|$]]|$1|1D|3|1E|5|1C|7|2Z|8|@$9|30|A|31|B|C]|$9|32|A|33|B|C]]|D|@]|E|$]]|$1|1F|3|-4|5|6|7|34|8|@]|D|@]|E|$]]|$1|1G|3|1H|5|6|7|35|8|@$9|36|A|37|B|C]|$9|38|A|39|B|C]]|D|@]|E|$]]|$1|1I|3|1J|5|6|7|3A|8|@]|D|@$9|3B|A|3C|1|3D]]|E|$]]|$1|1K|3|-4|5|6|7|3E|8|@]|D|@]|E|$]]]|1L|$1M|$5|1N|1O|1P|E|$1Q|1R]]|1S|$5|1N|1O|1P|E|$1Q|1T]]]]

See my answer to the question <a href="https://stackoverflow.com/questions/714063/importing-modules-from-parent-folder/33532002#33532002">Importing modules from parent folder</a> for related information, including why my answer doesn't use the unreliable <code>__file__</code> variable. This simple solution should be cross-compatible with different operating systems as the modules <code>os</code> and <code>inspect</code> come as part of Python.

First, you need to import parts of the inspect and os modules.



<pre><code>from inspect import getsourcefile
from os.path import abspath
</code></pre>

Next, use the following line anywhere else it's needed in your Python code:

<pre><code>abspath(getsourcefile(lambda:0))
</code></pre>

<h3>How it works:</h3>

From the built-in module <code>os</code> (description below), the <code>abspath</code> tool is imported. 

<blockquote>
 OS routines for Mac, NT, or Posix depending on what system we're on.
</blockquote>

Then <code>getsourcefile</code> (description below) is imported from the built-in module <code>inspect</code>. 

<blockquote>
 Get useful information from live Python objects.
</blockquote>

<ul>
<li><code>abspath(path)</code> returns the absolute/full version of a file path</li>
<li><code>getsourcefile(lambda:0)</code> somehow gets the internal source file of the lambda function object, so returns <code>'&lt;pyshell#nn&gt;'</code> in the Python shell or returns the file path of the Python code currently being executed.</li>
</ul>

Using <code>abspath</code> on the result of <code>getsourcefile(lambda:0)</code> should make sure that the file path generated is the full file path of the Python file. 
This explained solution was originally based on code from the answer at <a href="https://stackoverflow.com/questions/2632199/how-do-i-get-the-path-of-the-current-executed-file-in-python/18489147#18489147">How do I get the path of the current executed file in Python?</a>.

blocks|key|311804|text|您可以从pathlib模块使用Path：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|311805|from+pathlib+import+Path

#+...

Path(__file__)|code-block|syntax|javascript|311806|您可以使用call+to+parent在path中进一步操作：|311807|Path(__file__).parent|311808|entityMap^0|4|7|F|4|0|0|D|6|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

You can use <code>Path</code> from the <code>pathlib</code> module:

<pre><code>from pathlib import Path

# ...

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

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

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

blocks|key|3050607|text|这个解决方案即使在可执行文件中也是健壮的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3050608|import+inspect,+os.path

filename+=+inspect.getframeinfo(inspect.currentframe()).filename
path+++++=+os.path.dirname(os.path.abspath(filename))|code-block|syntax|javascript|3050609|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|$]]

This solution is robust even in executables:
<pre><code>import inspect, os.path

filename = inspect.getframeinfo(inspect.currentframe()).filename
path = os.path.dirname(os.path.abspath(filename))
</code></pre>

blocks|key|3050611|text|只需添加以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3050612|from+sys+import+*
path_to_current_file+=+sys.argv[0]
print(path_to_current_file)|code-block|syntax|javascript|3050613|或者：|3050614|from+sys+import+*
print(sys.argv[0])|3050615|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|$]]

Simply add the following:

<pre><code>from sys import *
path_to_current_file = sys.argv[0]
print(path_to_current_file)
</code></pre>

Or:

<pre><code>from sys import *
print(sys.argv[0])
</code></pre>

blocks|key|3050617|text|您只需调用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3050618|path+=+os.path.abspath(os.path.dirname(sys.argv[0]))|code-block|syntax|javascript|3050619|而不是：|3050620|path+=+os.path.dirname(os.path.abspath(sys.argv[0]))|3050621|abspath()提供sys.argv[0]的绝对路径(代码所在的文件名)，dirname()返回不带文件名的目录路径。|offset|length|style|CODE|3050622|entityMap^0|0|0|0|0|0|9|B|B|12|9|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

You have simply called:

<pre><code>path = os.path.abspath(os.path.dirname(sys.argv[0]))
</code></pre>

instead of:

<pre><code>path = os.path.dirname(os.path.abspath(sys.argv[0]))
</code></pre>

<code>abspath()</code> gives you the absolute path of <code>sys.argv[0]</code> (the filename your code is in) and <code>dirname()</code> returns the directory path without the filename.

blocks|key|3050624|text|如果代码来自某个文件，则可以获取其全名|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3050625|sys._getframe().f_code.co_filename|code-block|syntax|javascript|3050626|您还可以检索f_code.co_name形式的函数名|offset|length|style|CODE|3050627|entityMap^0|0|0|6|E|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If the code is coming from a file, you can get its full name

<pre><code>sys._getframe().f_code.co_filename
</code></pre>

You can also retrieve the function name as <code>f_code.co_name</code>

blocks|key|3050629|text|主要的想法是，有人会运行你的python代码，但你需要找到离python文件最近的文件夹。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3050630|我的解决方案是：|3050631|import+os
print(os.path.dirname(os.path.abspath(__file__)))|code-block|syntax|javascript|3050632|有了os.path.dirname(os.path.abspath(__file__))，你可以使用它来保存照片，输出文件，...etc|offset|length|style|CODE|3050633|entityMap^0|0|0|0|2|16|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The main idea is, somebody will run your python code, but you need to get the folder nearest the python file.
My solution is:
<pre><code>import os
print(os.path.dirname(os.path.abspath(__file__)))
</code></pre>
With
<code>os.path.dirname(os.path.abspath(__file__))</code>
You can use it with to save photos, output files, ...etc

blocks|key|3050635|text|import+os
current_file_path=os.path.dirname(os.path.realpath('__file__'))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3050636|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import os
current_file_path=os.path.dirname(os.path.realpath('__file__'))
</code></pre>

This may seem like a newbie question, but it is not. Some common approaches don't work in all cases:
<h1>sys.argv[0]</h1>
This means using <code>path = os.path.abspath(os.path.dirname(sys.argv[0]))</code>, but this does not work if you are running from another Python script in another directory, and this can happen in real life.
<h1>__file__</h1>
This means using <code>path = os.path.abspath(os.path.dirname(__file__))</code>, but I found that this doesn't work:
<ul>
<li><code>py2exe</code> doesn't have a <code>__file__</code> attribute, but there is <a href="http://www.py2exe.org/index.cgi/WhereAmI" rel="nofollow noreferrer">a workaround</a></li>
<li>When you run from <a href="https://en.wikipedia.org/wiki/IDLE" rel="nofollow noreferrer">IDLE</a> with <code>execute()</code> there is no <code>__file__</code> attribute</li>
<li><a href="https://en.wikipedia.org/wiki/Mac_OS_X_Snow_Leopard" rel="nofollow noreferrer">Mac OS X v10.6</a> (Snow Leopard) where I get <code>NameError: global name '__file__' is not defined</code></li>
</ul>
Related questions with incomplete answers:
<ul>
<li><a href="https://stackoverflow.com/questions/1296501/python-find-path-to-file-being-run">Find path to currently running file</a></li>
<li><a href="https://stackoverflow.com/questions/1483827/python-path-to-current-file-depends-on-how-i-execute-the-program">Path to current file depends on how I execute the program</a></li>
<li><a href="https://stackoverflow.com/questions/2259503/how-to-know-the-path-of-the-running-script-in-python">How can I know the path of the running script in Python?</a></li>
<li><a href="https://stackoverflow.com/questions/509742/python-chdir-to-dir-the-py-script-is-in">Change directory to the directory of a Python script</a></li>
</ul>
I'm looking for a generic solution, one that would work in all above use cases.
<h3>Update</h3>
Here is the result of a testcase:
<h3>Output of <code>python a.py</code> (on Windows)</h3>
<pre><code>a.py: __file__= a.py
a.py: os.getcwd()= C:\zzz

b.py: sys.argv[0]= a.py
b.py: __file__= a.py
b.py: os.getcwd()= C:\zzz
</code></pre>
<h3>a.py</h3>
<pre><code>#! /usr/bin/env python
import os, sys

print &quot;a.py: sys.argv[0]=&quot;, sys.argv[0]
print &quot;a.py: __file__=&quot;, __file__
print &quot;a.py: os.getcwd()=&quot;, os.getcwd()
print

execfile(&quot;subdir/b.py&quot;)
</code></pre>
<h3>File subdir/b.py</h3>
<pre><code>#! /usr/bin/env python
import os, sys

print &quot;b.py: sys.argv[0]=&quot;, sys.argv[0]
print &quot;b.py: __file__=&quot;, __file__
print &quot;b.py: os.getcwd()=&quot;, os.getcwd()
print
</code></pre>
<h3>tree</h3>
<pre><code>C:.
| a.py
\---subdir
 b.py
</code></pre>

How do I get the path of the current executed file in Python?

解释器

Windows

Python

Linux

这可能看起来像一个新手问题，但事实并非如此。一些常见的方法并不适用于所有情况：sys.argv这意味着使用path = os.path.abspath(os.path.dirname(sys.argv[0]))，但如果您从另一个目录中的另一个Python脚本运行，这将不起作用，这在现实生活中可能会发生。__file__...

问如何在Python中获取当前执行文件的路径？
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