blocks|key|294358|text|将会|type|unstyled|depth|inlineStyleRanges|entityRanges|data|294359|import+os
cwd+=+os.getcwd()|code-block|syntax|javascript|294360|做你想做的事?我不确定你所说的“当前脚本目录”到底是什么意思。对于您给出的用例，预期输出是什么？|294361|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|$]]

Would

<pre><code>import os
cwd = os.getcwd()
</code></pre>

do what you want? I'm not sure what exactly you mean by the "current script directory". What would the expected output be for the use cases you gave?

blocks|key|294363|text|os.path.dirname(os.path.abspath(__file__))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|294364|确实是你能得到的最好的。|unstyled|294365|使用exec/execfile执行脚本并不常见；通常情况下，您应该使用模块基础设施来加载脚本。如果必须使用这些方法，我建议在传递给脚本的globals中设置__file__，以便脚本可以读取该文件名。|offset|length|style|CODE|294366|没有其他方法可以在可执行代码中获取文件名:正如您所注意到的，CWD可能位于完全不同的位置。|294367|entityMap^0|0|0|2|4|7|8|1W|7|26|8|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|F|7|11|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|P|$]]

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

is indeed the best you're going to get.

It's unusual to be executing a script with <code>exec</code>/<code>execfile</code>; normally you should be using the module infrastructure to load scripts. If you must use these methods, I suggest setting <code>__file__</code> in the <code>globals</code> you pass to the script so it can read that filename.

There's no other way to get the filename in execed code: as you note, the CWD may be in a completely different place.

blocks|key|3236409|text|如果您真的想介绍通过execfile(...)调用脚本的情况，可以使用inspect模块来推断文件名(包括路径)。据我所知，这将适用于您列出的所有案例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3236410|filename+=+inspect.getframeinfo(inspect.currentframe()).filename
path+=+os.path.dirname(os.path.abspath(filename))|code-block|syntax|javascript|3236411|entityMap^0|A|D|Z|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

If you really want to cover the case that a script is called via <code>execfile(...)</code>, you can use the <code>inspect</code> module to deduce the filename (including the path). As far as I am aware, this will work for all cases you listed:

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

blocks|key|3236428|text|只需使用os.path.dirname(os.path.abspath(__file__))，并仔细检查使用exec的情况是否真的需要。如果你不能把你的脚本当做一个模块来使用，这可能是设计出问题的征兆。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3236429|请记住Python8的禅宗，如果您认为一个必须为exec工作的用例有很好的论据，那么请让我们了解有关问题背景的更多细节。|3236430|entityMap^0|4|16|1I|4|0|O|4|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

Just use <code>os.path.dirname(os.path.abspath(__file__))</code> and examine very carefully whether there is a real need for the case where <code>exec</code> is used. It could be a sign of troubled design if you are not able to use your script as a module. 

Keep in mind Zen of Python #8, and if you believe there is a good argument for a use-case where it must work for <code>exec</code>, then please let us know some more details about the background of the problem.

blocks|key|299461|text|首先..如果我们讨论的是注入匿名代码的方法，这里有几个缺失的用例。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|299462|code.compile_command()
code.interact()
imp.load_compiled()
imp.load_dynamic()
imp.load_module()
__builtin__.compile()
loading+C+compiled+shared+objects?+example:+_socket?)|code-block|syntax|javascript|299463|但是，真正的问题是，你的目标是什么--你是在试图实施某种安全措施吗？或者你只是对正在加载的东西感兴趣。|299464|如果您对security感兴趣，那么通过exec/execfile导入的文件名是无关紧要的-您应该使用rexec，它提供了以下功能：|offset|length|299465|299466|此模块包含RExec类，该类支持r_eval()、r_execfile()、r_exec()和r_import()方法，这些方法是标准Python函数eval()、execfile()以及exec和import语句的受限版本。在这个受限制的环境中执行的代码只能访问被认为是安全的模块和函数；您可以根据需要继承RExec，添加或删除功能。|blockquote|299467|299468|299469|然而，如果这更多的是一种学术追求..这里有几个愚蠢的方法，你也许可以深入挖掘一下……|299470|示例脚本：|299471|./deep.py|style|BOLD|299472|print+'+>>+level+1'
execfile('deeper.py')
print+'+<<+level+1'|299473|./deeper.py|299474|print+'\t+>>+level+2'
exec("import+sys;+sys.path.append('/tmp');+import+deepest")
print+'\t+<<+level+2'|299475|/tmp/deepest.py|299476|print+'\t\t+>>+level+3'
print+'\t\t\t+I+can+see+the+earths+core.'
print+'\t\t+<<+level+3'|299477|./codespy.py|299478|import+sys,+os

def+overseer(frame,+event,+arg):
++++print+"loaded(%25s)"+%25+os.path.abspath(frame.f_code.co_filename)

sys.settrace(overseer)
execfile("deep.py")
sys.exit(0)|299479|输出|299480|loaded(/Users/synthesizerpatel/deep.py)
>>+level+1
loaded(/Users/synthesizerpatel/deeper.py)
++++>>+level+2
loaded(/Users/synthesizerpatel/<string>)
loaded(/tmp/deepest.py)
++++++++>>+level+3
++++++++++++I+can+see+the+earths+core.
++++++++<<+level+3
++++<<+level+2
<<+level+1|299481|当然，这是一种资源密集型的方法，您将跟踪所有代码。效率不是很高。但是，我认为这是一种新颖的方法，因为即使你深入到巢中，它也会继续工作。您不能重写'eval‘。尽管您可以重写execfile()。|299482|注意，这种方法只覆盖exec/execfile，而不是“import”。对于更高级别的“模块”加载挂钩，你可以使用sys.path_hooks+(由PyMOTW提供)。|299483|这就是我脑海中所有的东西。|299484|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Security_through_obscurity|1|http://docs.python.org/library/rexec.html|2|http://blog.doughellmann.com/2009/11/pymotw-sys-part-7-modules-and-imports.html^0|0|0|0|4|8|0|1F|5|1|0|0|0|0|0|0|0|0|9|0|0|0|B|0|0|0|F|0|0|0|C|0|0|0|2|0|0|0|1L|E|2|0|0^^$0|@$1|2|3|4|5|6|7|20|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|21|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|22|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|23|8|@]|9|@$K|24|L|25|1|26]|$K|27|L|28|1|29]]|A|$]]|$1|M|3|-4|5|6|7|2A|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|2B|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|2C|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|2D|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|2E|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|2F|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|2G|8|@$K|2H|L|2I|Y|Z]]|9|@]|A|$]]|$1|10|3|11|5|D|7|2J|8|@]|9|@]|A|$E|F]]|$1|12|3|13|5|6|7|2K|8|@$K|2L|L|2M|Y|Z]]|9|@]|A|$]]|$1|14|3|15|5|D|7|2N|8|@]|9|@]|A|$E|F]]|$1|16|3|17|5|6|7|2O|8|@$K|2P|L|2Q|Y|Z]]|9|@]|A|$]]|$1|18|3|19|5|D|7|2R|8|@]|9|@]|A|$E|F]]|$1|1A|3|1B|5|6|7|2S|8|@$K|2T|L|2U|Y|Z]]|9|@]|A|$]]|$1|1C|3|1D|5|D|7|2V|8|@]|9|@]|A|$E|F]]|$1|1E|3|1F|5|6|7|2W|8|@$K|2X|L|2Y|Y|Z]]|9|@]|A|$]]|$1|1G|3|1H|5|D|7|2Z|8|@]|9|@]|A|$E|F]]|$1|1I|3|1J|5|6|7|30|8|@]|9|@]|A|$]]|$1|1K|3|1L|5|6|7|31|8|@]|9|@$K|32|L|33|1|34]]|A|$]]|$1|1M|3|1N|5|6|7|35|8|@]|9|@]|A|$]]|$1|1O|3|-4|5|6|7|36|8|@]|9|@]|A|$]]]|1P|$1Q|$5|1R|1S|1T|A|$1U|1V]]|1W|$5|1R|1S|1T|A|$1U|1X]]|1Y|$5|1R|1S|1T|A|$1U|1Z]]]]

First.. a couple missing use-cases here if we're talking about ways to inject anonymous code..

<pre><code>code.compile_command()
code.interact()
imp.load_compiled()
imp.load_dynamic()
imp.load_module()
__builtin__.compile()
loading C compiled shared objects? example: _socket?)
</code></pre>

But, the real question is, what is your goal - are you trying to enforce some sort of security? Or are you just interested in whats being loaded.

If you're interested in <a href="http://en.wikipedia.org/wiki/Security_through_obscurity" rel="nofollow">security</a>, the filename that is being imported via exec/execfile is inconsequential - you should use <a href="http://docs.python.org/library/rexec.html" rel="nofollow">rexec</a>, which offers the following:

<blockquote>
 This module contains the RExec class,
 which supports r_eval(), r_execfile(),
 r_exec(), and r_import() methods, which 
 are restricted versions of the standard
 Python functions eval(), execfile() and 
 the exec and import statements. Code 
 executed in this restricted environment
 will only have access to modules and 
 functions that are deemed safe; you can 
 subclass RExec add or remove capabilities as
 desired.
</blockquote>

However, if this is more of an academic pursuit.. here are a couple goofy approaches that you
might be able to dig a little deeper into.. 

Example scripts:

./deep.py

<pre><code>print ' &gt;&gt; level 1'
execfile('deeper.py')
print ' &lt;&lt; level 1'
</code></pre>

./deeper.py

<pre><code>print '\t &gt;&gt; level 2'
exec("import sys; sys.path.append('/tmp'); import deepest")
print '\t &lt;&lt; level 2'
</code></pre>

/tmp/deepest.py

<pre><code>print '\t\t &gt;&gt; level 3'
print '\t\t\t I can see the earths core.'
print '\t\t &lt;&lt; level 3'
</code></pre>

./codespy.py

<pre><code>import sys, os

def overseer(frame, event, arg):
 print "loaded(%s)" % os.path.abspath(frame.f_code.co_filename)

sys.settrace(overseer)
execfile("deep.py")
sys.exit(0)
</code></pre>

Output

<pre><code>loaded(/Users/synthesizerpatel/deep.py)
&gt;&gt; level 1
loaded(/Users/synthesizerpatel/deeper.py)
 &gt;&gt; level 2
loaded(/Users/synthesizerpatel/&lt;string&gt;)
loaded(/tmp/deepest.py)
 &gt;&gt; level 3
 I can see the earths core.
 &lt;&lt; level 3
 &lt;&lt; level 2
&lt;&lt; level 1
</code></pre>

Of course, this is a resource-intensive way to do it, you'd be tracing
all your code.. Not very efficient. But, I think it's a novel approach
since it continues to work even as you get deeper into the nest.
You can't override 'eval'. Although you can override execfile().

Note, this approach only coveres exec/execfile, not 'import'.
For higher level 'module' load hooking you might be able to use use
<a href="http://blog.doughellmann.com/2009/11/pymotw-sys-part-7-modules-and-imports.html" rel="nofollow">sys.path_hooks</a> (Write-up courtesy of PyMOTW).

Thats all I have off the top of my head.

blocks|key|299486|text|这是一个部分的解决方案，仍然比目前发布的所有解决方案都要好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|299487|import+sys,+os,+os.path,+inspect

#os.chdir("..")

if+'__file__'+not+in+locals():
++++__file__+=+inspect.getframeinfo(inspect.currentframe())[0]

print+os.path.dirname(os.path.abspath(__file__))|code-block|syntax|javascript|299488|现在，这将适用于所有调用，但如果有人使用chdir()更改当前目录，则此操作也将失败。|offset|length|style|CODE|299489|备注：|299490|299491|sys.argv[0]不会工作，如果您使用python+-c+"execfile('path-tester.py')"|unordered-list-item|299492|I执行脚本，将返回-c。https://gist.github.com/1385555发布了一个完整的测试，欢迎您对其进行改进。|299493|299494|entityMap|0|LINK|mutability|MUTABLE|url|https://gist.github.com/1385555^0|0|0|K|7|0|0|0|0|B|L|12|0|9|2|C|V|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@$I|16|J|17|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1A|8|@$I|1B|J|1C|K|L]|$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|S|3|T|5|R|7|1F|8|@$I|1G|J|1H|K|L]]|9|@$I|1I|J|1J|1|1K]]|A|$]]|$1|U|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12]]]]

Here is a partial solution, still better than all published ones so far.

<pre><code>import sys, os, os.path, inspect

#os.chdir("..")

if '__file__' not in locals():
 __file__ = inspect.getframeinfo(inspect.currentframe())[0]

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

Now this works will all calls but if someone use <code>chdir()</code> to change the current directory, this will also fail.

Notes: 

<ul>
<li><code>sys.argv[0]</code> is not going to work, will return <code>-c</code> if you execute the script with <code>python -c "execfile('path-tester.py')"</code></li>
<li>I published a complete test at <a href="https://gist.github.com/1385555" rel="nofollow">https://gist.github.com/1385555</a> and you are welcome to improve it.</li>
</ul>

blocks|key|296915|text|#!/usr/bin/env+python
import+inspect
import+os
import+sys

def+get_script_dir(follow_symlinks=True):
++++if+getattr(sys,+'frozen',+False):+#+py2exe,+PyInstaller,+cx_Freeze
++++++++path+=+os.path.abspath(sys.executable)
++++else:
++++++++path+=+inspect.getabsfile(get_script_dir)
++++if+follow_symlinks:
++++++++path+=+os.path.realpath(path)
++++return+os.path.dirname(path)

print(get_script_dir())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|296916|它适用于CPython，Jython，Pypy。如果使用execfile()执行脚本，则可以正常工作(sys.argv[0]和__file__+-based解决方案在这里将失败)。如果脚本在an+executable+zip+file+(/an+egg)中，它就会起作用。如果脚本是从压缩文件“导入”(PYTHONPATH=/path/to/library.zip+python+-mscript_to_run)的，那么它就可以工作；在本例中，它返回归档路径。如果脚本被编译成独立的可执行文件(sys.frozen)，它就会起作用。它适用于符号链接(realpath消除了符号链接)。它在交互式解释器中工作；在本例中，它返回当前的工作目录。|unstyled|offset|length|style|CODE|296917|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/5356563/4279^0|0|S|A|1F|B|1R|8|47|1I|6V|A|7N|8|2N|W|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@$G|U|H|V|I|J]|$G|W|H|X|I|J]|$G|Y|H|Z|I|J]|$G|10|H|11|I|J]|$G|12|H|13|I|J]|$G|14|H|15|I|J]]|9|@$G|16|H|17|1|18]]|A|$]]|$1|K|3|-4|5|F|7|19|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>#!/usr/bin/env python
import inspect
import os
import sys

def get_script_dir(follow_symlinks=True):
 if getattr(sys, 'frozen', False): # py2exe, PyInstaller, cx_Freeze
 path = os.path.abspath(sys.executable)
 else:
 path = inspect.getabsfile(get_script_dir)
 if follow_symlinks:
 path = os.path.realpath(path)
 return os.path.dirname(path)

print(get_script_dir())
</code></pre>

It works on CPython, Jython, Pypy. It works if the script is executed using <code>execfile()</code> (<code>sys.argv[0]</code> and <code>__file__</code> -based solutions would fail here). It works if the script is inside <a href="https://stackoverflow.com/a/5356563/4279">an executable zip file (/an egg)</a>. It works if the script is "imported" (<code>PYTHONPATH=/path/to/library.zip python -mscript_to_run</code>) from a zip file; it returns the archive path in this case. It works if the script is compiled into a standalone executable (<code>sys.frozen</code>). It works for symlinks (<code>realpath</code> eliminates symbolic links). It works in an interactive interpreter; it returns the current working directory in this case.

blocks|key|296919|text|在大多数情况下，这应该是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|296920|import+os,sys
dirname=os.path.dirname(os.path.realpath(sys.argv[0]))|code-block|syntax|javascript|296921|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 should work in most cases:

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

blocks|key|296934|text|希望这会有所帮助：-如果你从任何地方运行脚本/模块，你将能够访问__file__变量，这是一个代表脚本位置的模块变量。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|296935|另一方面，如果你使用解释器，你不能访问这个变量，你会得到一个名字NameError，如果你从其他地方运行文件，os.getcwd()会给你一个错误的目录。|296936|This解决方案应该在所有情况下都能满足您的需求：|296937|from+inspect+import+getsourcefile
from+os.path+import+abspath
abspath(getsourcefile(lambda:0))|code-block|syntax|javascript|296938|我还没有彻底测试它，但它解决了我的问题。|296939|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/18489147/7609849^0|W|8|0|W|9|1J|B|0|0|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|J|3|K|5|L|7|1A|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1B|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

Hopefully this helps:- 
If you run a script/module from anywhere you'll be able to access the <code>__file__</code> variable which is a module variable representing the location of the script. 

On the other hand, if you're using the interpreter you don't have access to that variable, where you'll get a name <code>NameError</code> and <code>os.getcwd()</code> will give you the incorrect directory if you're running the file from somewhere else. 

<a href="https://stackoverflow.com/a/18489147/7609849">This</a> solution should give you what you're looking for in all cases:

<pre><code>from inspect import getsourcefile
from os.path import abspath
abspath(getsourcefile(lambda:0))
</code></pre>

I haven't thoroughly tested it but it solved my problem.

blocks|key|3022081|text|在Python+3.4%2B中，您可以使用更简单的pathlib模块：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3022082|from+inspect+import+currentframe,+getframeinfo
from+pathlib+import+Path

filename+=+getframeinfo(currentframe()).filename
parent+=+Path(filename).resolve().parent|code-block|syntax|javascript|3022083|您还可以使用__file__完全避免使用inspect模块：|3022084|from+pathlib+import+Path
parent+=+Path(__file__).resolve().parent|3022085|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/pathlib.html^0|N|7|N|7|0|0|0|6|8|K|7|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

In Python 3.4+ you can use the simpler <a href="https://docs.python.org/3/library/pathlib.html" rel="noreferrer"><code>pathlib</code></a> module:
<pre><code>from inspect import currentframe, getframeinfo
from pathlib import Path

filename = getframeinfo(currentframe()).filename
parent = Path(filename).resolve().parent
</code></pre>
You can also use <code>__file__</code> to avoid the <code>inspect</code> module altogether:
<pre class="lang-py prettyprint-override"><code>from pathlib import Path
parent = Path(__file__).resolve().parent
</code></pre>

blocks|key|297572|text|os.path...方法是Python2中的“已经完成的事情”。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|297573|在Python+3中，您可以找到脚本目录，如下所示：|297574|from+pathlib+import+Path
script_path+=+Path(__file__).parent|code-block|syntax|javascript|297575|entityMap^0|0|A|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

The <code>os.path...</code> approach was the 'done thing' in Python 2.
In Python 3, you can find directory of script as follows:
<pre><code>from pathlib import Path
script_path = Path(__file__).parent
</code></pre>

blocks|key|297071|text|注意:这个答案现在是一个包|type|unstyled|depth|inlineStyleRanges|entityRanges|data|297072|https://github.com/heetbeet/locate|offset|length|297073|$+pip+install+locate

$+python
>>>+from+locate+import+this_dir
>>>+print(this_dir())
C:/Users/simon|code-block|syntax|javascript|297074|对于.py脚本以及交互式用法：|style|CODE|297075|我经常使用我的脚本所在的目录(用于访问与脚本一起存储的文件)，但出于调试目的，我也经常在交互式shell中运行这些脚本。我将__dirpath__定义为：|297076|运行或导入.py文件时的|297077|，即文件的基目录。这始终是正确的路径。|unordered-list-item|297078|运行.ipyn笔记本时，当前工作目录。这始终是正确的路径，因为Jupyter将工作目录设置为在REPL中运行的.ipynb基本directory.|297079|When，即当前工作目录。嗯，当代码从文件中分离出来时，实际的“正确路径”是什么？相反，在调用REPL之前，您有责任更改为“正确的路径”。|297080|297081|Python+3.4+(及更高版本)：|297082|from+pathlib+import+Path
__dirpath__+=+Path(globals().get("__file__",+"./_")).absolute().parent|297083|Python+2(及更高版本)：|297084|import+os
__dirpath__+=+os.path.dirname(os.path.abspath(globals().get("__file__",+"./_")))|297085|解释：|297086|297087|globals()将所有全局变量作为dictionary.|297088|.get("__file__",+"./_")返回，如果键"__file__"存在于globals()中，则返回该值；否则，它将返回提供的默认值"./_".|297089|The其余代码只是将__file__+(或"./_")展开为绝对文件路径，然后返回文件路径的基目录。|297090|297091|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|Y|0|0|0|2|3|0|1Q|B|0|5|3|0|0|2|5|1J|6|0|0|0|0|0|0|0|0|0|0|9|0|0|N|T|A|16|9|20|5|0|A|8|L|5|0|0^^$0|@$1|2|3|4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1Q|8|@]|9|@$D|1R|E|1S|1|1T]]|A|$]]|$1|F|3|G|5|H|7|1U|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|1V|8|@$D|1W|E|1X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1Y|8|@$D|1Z|E|20|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|21|8|@$D|22|E|23|M|N]]|9|@]|A|$]]|$1|S|3|T|5|U|7|24|8|@]|9|@]|A|$]]|$1|V|3|W|5|U|7|25|8|@$D|26|E|27|M|N]|$D|28|E|29|M|N]]|9|@]|A|$]]|$1|X|3|Y|5|U|7|2A|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|2B|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|2C|8|@]|9|@]|A|$]]|$1|12|3|13|5|H|7|2D|8|@]|9|@]|A|$I|J]]|$1|14|3|15|5|6|7|2E|8|@]|9|@]|A|$]]|$1|16|3|17|5|H|7|2F|8|@]|9|@]|A|$I|J]]|$1|18|3|19|5|6|7|2G|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|2H|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|U|7|2I|8|@$D|2J|E|2K|M|N]]|9|@]|A|$]]|$1|1D|3|1E|5|U|7|2L|8|@$D|2M|E|2N|M|N]|$D|2O|E|2P|M|N]|$D|2Q|E|2R|M|N]|$D|2S|E|2T|M|N]]|9|@]|A|$]]|$1|1F|3|1G|5|U|7|2U|8|@$D|2V|E|2W|M|N]|$D|2X|E|2Y|M|N]]|9|@]|A|$]]|$1|1H|3|-4|5|6|7|2Z|8|@]|9|@]|A|$]]|$1|1I|3|-4|5|6|7|30|8|@]|9|@]|A|$]]]|1J|$1K|$5|1L|1M|1N|A|$1O|C]]]]

<h2>Note: this answer is now a package</h2>
<a href="https://github.com/heetbeet/locate" rel="nofollow noreferrer">https://github.com/heetbeet/locate</a>
<pre><code>$ pip install locate

$ python
&gt;&gt;&gt; from locate import this_dir
&gt;&gt;&gt; print(this_dir())
C:/Users/simon
</code></pre>
<h2>For <code>.py</code> scripts as well as interactive usage:</h2>
I frequently use the directory of my scripts (for accessing files stored along side them), but I also frequently run these scripts in an interactive shell for debugging purposes. I define <code>__dirpath__</code> as:
<ul>
<li>When running or importing a <code>.py</code> file, the file's base directory. This is always the correct path.</li>
<li>When running an <code>.ipyn</code> notebook, the current working directory. This is always the correct path, since Jupyter sets the working directory as the <code>.ipynb</code> base directory.</li>
<li>When running in a REPL, the current working directory. Hmm, what is the actual &quot;correct path&quot; when the code is detached from a file? Rather, make it your responsibility to change into the &quot;correct path&quot; before invoking the REPL.</li>
</ul>
<h3>Python 3.4 (and above):</h3>
<pre><code>from pathlib import Path
__dirpath__ = Path(globals().get(&quot;__file__&quot;, &quot;./_&quot;)).absolute().parent
</code></pre>
<h3>Python 2 (and above):</h3>
<pre><code>import os
__dirpath__ = os.path.dirname(os.path.abspath(globals().get(&quot;__file__&quot;, &quot;./_&quot;)))
</code></pre>
<h2>Explanation:</h2>
<ul>
<li><code>globals()</code> returns all the global variables as a dictionary.</li>
<li><code>.get(&quot;__file__&quot;, &quot;./_&quot;)</code> returns the value from the key <code>&quot;__file__&quot;</code> if it exists in <code>globals()</code>, otherwise it returns the provided default value <code>&quot;./_&quot;</code>.</li>
<li>The rest of the code just expands <code>__file__</code> (or <code>&quot;./_&quot;</code>) into an absolute filepath, and then returns the filepath's base directory.</li>
</ul>

blocks|key|3022230|text|如果__file__可用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3022231|#+--+script1.py+--
import+os
file_path+=+os.path.abspath(__file__)
print(os.path.dirname(file_path))|code-block|syntax|javascript|3022232|对于那些我们希望能够从解释器运行命令或获取运行脚本的位置的路径：|3022233|#+--+script2.py+--
import+os
print(os.path.abspath(''))|3022234|这在解释器中起作用。但当在脚本中运行(或导入)时，它会提供运行脚本的位置的路径，而不是包含带有打印的脚本的目录的路径。|3022235|示例：|3022236|如果您的目录结构是|3022237|test_dir+(in+the+home+dir)
├──+main.py
└──+test_subdir
++++├──+script1.py
++++└──+script2.py|3022238|使用|3022239|#+--+main.py+--
import+script1.py
import+script2.py|3022240|输出为：|3022241|~/test_dir/test_subdir
~/test_dir|3022242|entityMap^0|2|8|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1D|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1E|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|10|3|11|5|6|7|1I|8|@]|D|@]|E|$]]|$1|12|3|13|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|14|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|15|$]]

If <code>__file__</code> is available:
<pre class="lang-py prettyprint-override"><code># -- script1.py --
import os
file_path = os.path.abspath(__file__)
print(os.path.dirname(file_path))
</code></pre>
For those we want to be able to run the command from the interpreter or get the path of the place you're running the script from:
<pre class="lang-py prettyprint-override"><code># -- script2.py --
import os
print(os.path.abspath(''))
</code></pre>
This works from the interpreter.
But when run in a script (or imported) it gives the path of the place where
you ran the script from, not the path of directory containing
the script with the print.
<h3>Example:</h3>
If your directory structure is
<pre><code>test_dir (in the home dir)
├── main.py
└── test_subdir
 ├── script1.py
 └── script2.py
</code></pre>
with
<pre class="lang-py prettyprint-override"><code># -- main.py --
import script1.py
import script2.py
</code></pre>
The output is:
<pre><code>~/test_dir/test_subdir
~/test_dir
</code></pre>

blocks|key|3236754|text|print(__import__("pathlib").Path(__file__).parent)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3236755|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>print(__import__(&quot;pathlib&quot;).Path(__file__).parent)
</code></pre>

blocks|key|299689|text|要获取包含当前脚本的目录的绝对路径，可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|299690|from+pathlib+import+Path
absDir+=+Path(__file__).parent.resolve()|code-block|syntax|javascript|299691|请注意，.resolve()调用是必需的，因为它使路径成为绝对路径。如果没有resolve()，您将获得类似于'.'的东西。|offset|length|style|CODE|299692|这个解决方案使用pathlib，它是Python的stdlib的一部分，从v3.4+(2014)开始。与使用os的其他解决方案相比，这是更可取的。|299693|官方的pathlib文档有一个有用的表，它将旧的os函数映射到新的函数：https://docs.python.org/3/library/pathlib.html#correspondence-to-tools-in-the-os-module|299694|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/pathlib.html|1|https://docs.python.org/3/library/pathlib.html#correspondence-to-tools-in-the-os-module^0|0|0|4|A|12|9|1J|3|0|8|7|1I|2|8|7|0|0|3|7|O|2|10|2F|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@$I|1A|J|1B|K|L]|$I|1C|J|1D|K|L]]|9|@$I|1E|J|1F|1|1G]]|A|$]]|$1|O|3|P|5|6|7|1H|8|@$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]]|9|@$I|1M|J|1N|1|1O]]|A|$]]|$1|Q|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

To get the absolute path to the directory containing the current script you can use:
<pre class="lang-py prettyprint-override"><code>from pathlib import Path
absDir = Path(__file__).parent.resolve()
</code></pre>
Please note the <code>.resolve()</code> call is required, because that is the one making the path absolute. Without <code>resolve()</code>, you would obtain something like <code>'.'</code>.
This solution uses <a href="https://docs.python.org/3/library/pathlib.html" rel="nofollow noreferrer"><code>pathlib</code></a>, which is part of Python's stdlib since v3.4 (2014). This is preferrable compared to other solutions using <code>os</code>.
The official <code>pathlib</code> documentation has a useful table mapping the old <code>os</code> functions to the new ones: <a href="https://docs.python.org/3/library/pathlib.html#correspondence-to-tools-in-the-os-module" rel="nofollow noreferrer">https://docs.python.org/3/library/pathlib.html#correspondence-to-tools-in-the-os-module</a>

blocks|key|297684|text|由于前面的答案需要你导入一些模块，我想我会写一个不需要的答案。如果你不想导入任何东西，请使用下面的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|297685|dir+=+'/'.join(__file__.split('/')[:-1])
print(dir)|code-block|syntax|javascript|297686|如果脚本在/path/to/script.py上，那么这将打印/path/to。请注意，这将在终端上抛出错误，因为没有执行任何文件。这基本上解析了__file__中的目录，删除了目录的最后一部分。在这种情况下，删除/script.py以生成输出/path/to。|offset|length|style|CODE|297687|entityMap^0|0|0|5|I|V|8|21|8|2Z|A|3E|8|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]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$]]

As previous answers require you to import some module, I thought that I would write one answer that doesn't. Use the code below if you don't want to import anything.
<pre class="lang-py prettyprint-override"><code>dir = '/'.join(__file__.split('/')[:-1])
print(dir)
</code></pre>
If the script is on <code>/path/to/script.py</code> then this would print <code>/path/to</code>. Note that this will throw error on terminal as no file is executed. This basically parse the directory from <code>__file__</code> removing the last part of it. In this case <code>/script.py</code> is removed to produce the output <code>/path/to</code>.

I would like to see what is the best way to determine the current script directory in Python.
I discovered that, due to the many ways of calling Python code, it is hard to find a good solution.
Here are some problems:
<ul>
<li><code>__file__</code> is not defined if the script is executed with <code>exec</code>, <code>execfile</code></li>
<li><code>__module__</code> is defined only in modules</li>
</ul>
Use cases:
<ul>
<li><code>./myfile.py</code></li>
<li><code>python myfile.py</code></li>
<li><code>./somedir/myfile.py</code></li>
<li><code>python somedir/myfile.py</code></li>
<li><code>execfile('myfile.py')</code> (from another script, that can be located in another directory and that can have another current directory.</li>
</ul>
I know that there is no perfect solution, but I'm looking for the best approach that solves most of the cases.
The most used approach is <code>os.path.dirname(os.path.abspath(__file__))</code> but this really doesn't work if you execute the script from another one with <code>exec()</code>.
<h2>Warning</h2>
Any solution that uses current directory will fail, this can be different based on the way the script is called or it can be changed inside the running script.

How do you properly determine the current script directory?

解释器

Python

我想看看在Python中确定当前脚本目录的最佳方法是什么。我发现，由于调用Python代码的方式很多，很难找到一个好的解决方案。以下是一些问题：如果使用模块执行脚本，则不定义__file__；如果仅在execfile__module__ 中定义模块，则不定义exec使用案例：./myfile.pypython myfi...

问如何正确确定当前脚本目录？
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问如何正确确定当前脚本目录？
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