python NameError:未定义全局名称"__file__"

Question

当我在python 2.7中运行此代码时,我收到此错误:

Traceback (most recent call last):
File "C:\Python26\Lib\site-packages\pyutilib.subprocess-3.5.4\setup.py", line 30, in <module>
    long_description = read('README.txt'),
  File "C:\Python26\Lib\site-packages\pyutilib.subprocess-3.5.4\setup.py", line 19, in read
    return open(os.path.join(os.path.dirname(__file__), *rnames)).read()
NameError: global name '__file__' is not defined

代码是:

import os
from setuptools import setup


def read(*rnames):
    return open(os.path.join(os.path.dirname(__file__), *rnames)).read()


setup(name="pyutilib.subprocess",
    version='3.5.4',
    maintainer='William E. Hart',
    maintainer_email='wehart@sandia.gov',
    url = 'https://software.sandia.gov/svn/public/pyutilib/pyutilib.subprocess',
    license = 'BSD',
    platforms = ["any"],
    description = 'PyUtilib utilites for managing subprocesses.',
    long_description = read('README.txt'),
    classifiers = [
        'Development Status :: 4 - Beta',
        'Intended Audience :: End Users/Desktop',
        'License :: OSI Approved :: BSD License',
        'Natural Language :: English',
        'Operating System :: Microsoft :: Windows',
        'Operating System :: Unix',
        'Programming Language :: Python',
        'Programming Language :: Unix Shell',
        'Topic :: Scientific/Engineering :: Mathematics',
        'Topic :: Software Development :: Libraries :: Python Modules'],
      packages=['pyutilib', 'pyutilib.subprocess', 'pyutilib.subprocess.tests'],
      keywords=['utility'],
      namespace_packages=['pyutilib'],
      install_requires=['pyutilib.common', 'pyutilib.services']
      )

Answer 1

当您os.path.join(os.path.dirname(__file__))在python交互式shell中追加此行时,会出现此错误.

Python Shell不检测当前文件路径__file__,它与filepath您添加此行的路径相关

所以,你应该写这条线os.path.join(os.path.dirname(__file__))在file.py.然后运行python file.py,它的工作原因是它需要你的文件路径.

Answer 2

我遇到了与PyInstaller和Py2exe相同的问题,所以我在cx-freeze中遇到了常见问题解答.

从控制台或作为应用程序使用脚本时,下面的函数将为您提供"执行路径",而不是"实际文件路径":

print(os.getcwd())
print(sys.argv[0])
print(os.path.dirname(os.path.realpath('__file__')))

资料来源:http:
//cx-freeze.readthedocs.org/en/latest/faq.html

你的旧行(初步问题):

def read(*rnames):
return open(os.path.join(os.path.dirname(__file__), *rnames)).read()

用以下代码替换您的代码行.

def find_data_file(filename):
    if getattr(sys, 'frozen', False):
        # The application is frozen
        datadir = os.path.dirname(sys.executable)
    else:
        # The application is not frozen
        # Change this bit to match where you store your data files:
        datadir = os.path.dirname(__file__)

    return os.path.join(datadir, filename)

使用上面的代码,您可以将应用程序添加到操作系统的路径中,您可以在任何地方执行它,而不会出现您的应用程序无法找到它的数据/配置文件的问题.

用python测试:

• 3.3.4
• 2.7.13

Answer 3

如果您在 .py 文件中使用代码：使用 os.path.abspath(__file__) 如果您直接在脚本上或在 Jupyter Notebooks 中使用代码：将文件放在双引号内。 os.path.abspath("__file__")

Answer 4

你在使用互动翻译吗？您可以使用

sys.argv[0]

您应该阅读:如何在Python中获取当前执行文件的路径？

Answer 5

我通过将文件视为字符串来解决它,即放入"__file__"(与引号一起!)而不是__file__

这对我来说很好:

wk_dir = os.path.dirname(os.path.realpath('__file__'))

Answer 6

如果您要查找的只是获取当前工作目录os.getcwd(),os.path.dirname(__file__)那么只要您没有更改代码中其他位置的工作目录,就可以获得相同的功能. os.getcwd()也可以在交互模式下工作.

因此 os.path.join(os.path.dirname(__file__)) 变得 os.path.join(os.getcwd())

Answer 7

如下更改您的代码！这个对我有用。`

os.path.dirname(os.path.abspath("__file__"))

Answer 8

如果你从 python shell 运行命令，你会得到这个：

>>> __file__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name '__file__' is not defined

您需要直接执行文件，将其作为参数传递给python命令：

$ python somefile.py

在你的情况下，它真的应该是 python setup.py install

Answer 9

I've run into cases where __file__ doesn't work as expected. But the following hasn't failed me so far:

import inspect
src_file_path = inspect.getfile(lambda: None)