subprocess:意外关键字参数capture_output

Question

当执行Python文档中subprocess.run()给出的时,我得到一个TypeError:

>>> import subprocess
>>> subprocess.run(["ls", "-l", "/dev/null"], capture_output=True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python3.6/subprocess.py", line 403, in run
    with Popen(*popenargs, **kwargs) as process:
TypeError: __init__() got an unexpected keyword argument 'capture_output'

我正在运行Python 3.6.6:

$ python3 --version
Python 3.6.6

Answer 1

你检查错误的文档中,蟒蛇,3.6这个参数并不会存在,因为可以在中找到文件(您选择左上角的版本):

subprocess.run(args, *, stdin=None, input=None, stdout=None, stderr=None,
               shell=False, cwd=None, timeout=None, check=False, encoding=None,
               errors=None, env=None)

但是,你可以通过设置stdout和stderr来轻松地"模拟"这个PIPE:

from subprocess import PIPE

subprocess.run(["ls", "-l", "/dev/null"], stdout=PIPE, stderr=PIPE)

实际上,如果我们查看python-3.7版本的源代码,其中介绍了该功能,我们在源代码[GitHub]中看到:

if capture_output:
    if ('stdout' in kwargs) or ('stderr' in kwargs):
        raise ValueError('stdout and stderr arguments may not be used '
                         'with capture_output.')
    kwargs['stdout'] = PIPE
    kwargs['stderr'] = PIPE

Answer 2

最简单的方法是使用subprocess.check_output函数：

import subprocess
subprocess.check_output(["ls", "-l", "/dev/null"])