Windows subprocess.Popen a batch file without shell=True [英] Windows subprocess.Popen a batch file without shell=True

问题描述

我有一个运行 lessc 的函数(用 npm install -g less 安装):

<预><代码>>>>导入子流程>>>subprocess.Popen(['lessc'])回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中文件C:\Python27\lib\subprocess.py"，第 679 行，在 __init__ 中错误读取，错误写入)文件C:\Python27\lib\subprocess.py"，第 896 行，在 _execute_child启动信息)WindowsError: [错误 2] 系统找不到指定的文件

不幸的是，除非我添加 shell=True，否则它不起作用:

<预><代码>>>>subprocess.Popen(['lessc'], shell=True)<subprocess.Popen 对象在 0x01F619D0>

如何在不使用 shell=True 的情况下使 lessc 运行?

解决方案

来自 https://docs.python.org/3/library/subprocess.html#subprocess.Popen 和 https://docs.python.org/2/library/subprocess.html#subprocess.Popen:

<块引用>

您不需要 shell=True 来运行批处理文件或基于控制台的可执行文件.

已经@JBernardo 引用.

那么，让我们试试:

lessc 实际告诉我们的地方

C:\Users\myname\AppData\Roaming\npm\lesscC:\Users\myname\AppData\Roaming\npm\lessc.cmd

也就是说，要执行的文件是lessc.cmd，而不是一些.bat 文件.事实上:

<预><代码>>>>导入子流程>>>subprocess.Popen([r'C:\Users\myname\AppData\Roaming\npm\lessc.cmd'])<subprocess.Popen 对象在 0x035BA070>>>>lessc:没有输入文件用法:lessc [option option=parameter ...] <source>[目的地]

因此，如果您指定完整路径，这确实有效.我认为当你有 这种体验时，有一个错字.可能你写的是 .bat 而不是 .cmd?

<小时>

如果你不想把lessc的完整路径打到你的脚本中，你可以自己烘焙一个where:

导入平台导入操作系统定义在哪里(文件名):# 灵感来自 http://nedbatchelder.com/code/utilities/wh.py# 另见:http://stackoverflow.com/questions/11210104/path_sep = ":" if platform.system() == "Linux" else ";"path_ext = [''] 如果 platform.system() == "Linux" 或 '.'在 file_name else os.environ["PATHEXT"].split(path_sep)对于 os.environ["PATH"].split(path_sep) 中的 d:对于路径_ext 中的 e:file_path = os.path.join(d, file_name + e)如果 os.path.exists(file_path):返回文件路径引发异常(文件名+未找到")

然后你可以写:

导入子流程subprocess.Popen([where('lessc')])

I have a function that runs lessc (installed with npm install -g less):

>>> import subprocess
>>> subprocess.Popen(['lessc'])
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python27\lib\subprocess.py", line 679, in __init__
    errread, errwrite)
  File "C:\Python27\lib\subprocess.py", line 896, in _execute_child
    startupinfo)
WindowsError: [Error 2] The system cannot find the file specified

Unfortunately, it doesn't work unless I add shell=True:

>>> subprocess.Popen(['lessc'], shell=True)
<subprocess.Popen object at 0x01F619D0>

What can I do to make lessc run without using shell=True?

解决方案

From both https://docs.python.org/3/library/subprocess.html#subprocess.Popen and https://docs.python.org/2/library/subprocess.html#subprocess.Popen:

You do not need shell=True to run a batch file or console-based executable.

as already cited by @JBernardo.

So, lets try:

where lessc actually tells

C:\Users\myname\AppData\Roaming\npm\lessc
C:\Users\myname\AppData\Roaming\npm\lessc.cmd

That means, the file to execute is lessc.cmd, not some .bat file. And indeed:

>>> import subprocess
>>> subprocess.Popen([r'C:\Users\myname\AppData\Roaming\npm\lessc.cmd'])
<subprocess.Popen object at 0x035BA070>
>>> lessc: no input files

usage: lessc [option option=parameter ...] <source> [destination]

So, this does work if you specify the full path. I assume there was a typo involved when you had this experience. May be you wrote .bat instead of .cmd?

---

If you don't want to patch the full path of lessc into your script, you can bake yourself a where:

import plaform
import os

def where(file_name):
    # inspired by http://nedbatchelder.com/code/utilities/wh.py
    # see also: http://stackoverflow.com/questions/11210104/
    path_sep = ":" if platform.system() == "Linux" else ";"
    path_ext = [''] if platform.system() == "Linux" or '.' in file_name else os.environ["PATHEXT"].split(path_sep)
    for d in os.environ["PATH"].split(path_sep):
        for e in path_ext:
            file_path = os.path.join(d, file_name + e)
            if os.path.exists(file_path):
                return file_path
    raise Exception(file_name + " not found")

Then you can write:

import subprocess
subprocess.Popen([where('lessc')])

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