如果你正在做一个不寻常的导入(例如，它是一个选项文件)，尝试:

import inspect
print (inspect.getfile(inspect.currentframe()))

注意，这将返回文件的绝对路径。

对于现代Python版本(3.4+)，Path(__file__).name应该更习惯。同时,路径(__file__)。Stem提供了不带.py扩展名的脚本名称。

由于OP要求当前脚本文件的名称，我更喜欢

import os
os.path.split(sys.argv[0])[1]

如果你正在做一个不寻常的导入(例如，它是一个选项文件)，尝试:

import inspect
print (inspect.getfile(inspect.currentframe()))

注意，这将返回文件的绝对路径。

为了完整起见，我认为有必要总结各种可能的结果，并为每种结果的确切行为提供参考。

答案由四个部分组成:

返回当前执行脚本的完整路径的不同方法的列表。 关于处理相对路径的警告。 关于符号链接处理的建议。 说明了一些方法，这些方法可用于从完整的文件路径中提取实际文件名(带或不带后缀)。

提取完整的文件路径

__file__ is the currently executing file, as detailed in the official documentation: __file__ is the pathname of the file from which the module was loaded, if it was loaded from a file. The __file__ attribute may be missing for certain types of modules, such as C modules that are statically linked into the interpreter; for extension modules loaded dynamically from a shared library, it is the pathname of the shared library file. From Python3.4 onwards, per issue 18416, __file__ is always an absolute path, unless the currently executing file is a script that has been executed directly (not via the interpreter with the -m command line option) using a relative path. __main__.__file__ (requires importing __main__) simply accesses the aforementioned __file__ attribute of the main module, e.g. of the script that was invoked from the command line. From Python3.9 onwards, per issue 20443, the __file__ attribute of the __main__ module became an absolute path, rather than a relative path. sys.argv[0] (requires importing sys) is the script name that was invoked from the command line, and might be an absolute path, as detailed in the official documentation: argv[0] is the script name (it is operating system dependent whether this is a full pathname or not). If the command was executed using the -c command line option to the interpreter, argv[0] is set to the string '-c'. If no script name was passed to the Python interpreter, argv[0] is the empty string. As mentioned in another answer to this question, Python scripts that were converted into stand-alone executable programs via tools such as py2exe or PyInstaller might not display the desired result when using this approach (i.e. sys.argv[0] would hold the name of the executable rather than the name of the main Python file within that executable). If none of the aforementioned options seem to work, probably due to an atypical execution process or an irregular import operation, the inspect module might prove useful, as suggested in another answer to this question: import inspect source_file_path = inspect.getfile(inspect.currentframe()) However, inspect.currentframe() would raise an exception when running in an implementation without Python stack frame. Note that inspect.getfile(...) is preferred over inspect.getsourcefile(...) because the latter raises a TypeError exception when it can determine only a binary file, not the corresponding source file (see also this answer to another question). From Python3.6 onwards, and as detailed in another answer to this question, it's possible to install an external open source library, lib_programname, which is tailored to provide a complete solution to this problem. This library iterates through all of the approaches listed above until a valid path is returned. If all of them fail, it raises an exception. It also tries to address various pitfalls, such as invocations via the pytest framework or the pydoc module. import lib_programname # this returns the fully resolved path to the launched python program path_to_program = lib_programname.get_path_executed_script() # type: pathlib.Path

处理相对路径

当处理一个恰好返回相对路径的方法时，可能很容易调用各种路径操作函数，例如os.path.abspath(…)或os.path.realpath(…)，以便提取完整或真实的路径。

但是，这些方法依赖于当前路径来获得完整路径。因此，如果程序首先更改当前工作目录，例如通过os.chdir(…)，然后才调用这些方法，则它们将返回错误的路径。

处理符号链接

如果当前脚本是一个符号链接，那么上述所有操作都将返回符号链接的路径，而不是实际文件的路径，为了提取后者，应该调用os.path.realpath(…)。

提取实际文件名的进一步操作

Os.path.basename(…)可能会被调用在上述任何一个上以提取实际的文件名，os.path.splitext(…)可能会被调用在实际的文件名上以截断其后缀，如os.path.splitext(Os.path.basename(…))。

从Python 3.4开始，根据PEP 428, pathlib模块的PurePath类也可以用于上述任何一个。具体来说，pathlib.PurePath(…).name提取实际的文件名和pathlib.PurePath(…)。Stem提取不带后缀的实际文件名。

所有的答案都很棒，但也有一些你第一眼可能看不出来的问题。

让我们定义我们想要的-我们想要执行的脚本的名称，而不是当前模块的名称-因此__file__只在已执行的脚本中使用，而不是在导入的模块中使用时才有效。 sys。Argv也是有问题的——如果你的程序是由pytest调用的呢?还是pydoc runner ?或者它是由uwsgi调用的?

还有第三种获取脚本名称的方法，我在答案中没有看到-你可以检查堆栈。

另一个问题是，你(或其他程序)可以篡改sys. exe。Argv和__main__。__file__ -它可能存在，也可能不存在。它可能成立，也可能不成立。至少您可以检查脚本(期望的结果)是否存在!

lib_programname库就是这样做的:

检查__main__是否存在 检查__main__。存在__file__ 给出__main__。__file__一个有效的结果(该脚本存在吗?) 如果不是:检查sys.argv: 系统中是否存在pytest, docrunner等。argv吗?——>如果是，忽略它 我们能得到一个有效的结果吗? 如果不是:检查堆栈，并尽可能从那里得到结果 如果堆栈也没有给出有效的结果，则抛出异常。

通过这种方式，我的解决方案到目前为止使用setup.py test, uwsgi, pytest, pycharm pytest, pycharm docrunner (doctest)， dreampie, eclipse

Dough Hellman也有一篇关于这个问题的不错的博客文章，“从Python中确定进程的名称”。

顺便说一句，它将在python 3.9中再次改变:主模块的文件属性变成了绝对路径，而不是相对路径。在当前目录被os.chdir()更改后，这些路径仍然有效

所以我宁愿照顾一个小模块，而不是浏览我的代码库，如果它应该改变一些…

免责声明:我是lib_programname库的作者。