相关疑难解决方法(0)

在安装了32位python 2.7的64位系统中,我尝试执行以下操作:

import subprocess
p = subprocess.call('dir', shell=True)
print p

但这给了我:

Traceback (most recent call last):
  File "test.py", line 2, in <module>
    p = subprocess.call('dir', shell=True)
  File "C:\Python27\lib\subprocess.py", line 522, in call
    return Popen(*popenargs, **kwargs).wait()
  File "C:\Python27\lib\subprocess.py", line 709, in __init__
    errread, errwrite)
  File "C:\Python27\lib\subprocess.py", line 957, in _execute_child
    startupinfo)
  WindowsError: [Error 2] The system cannot find the file specified

如果我在终端做...

dir

...当然会打印当前文件夹内容.

我试图将shell参数更改为shell = False.

编辑:其实我不能在路径上调用任何可执行文件subprocess.call().该声明p = subprocess.call('dir', shell=True)在另一台机器上工作正常,我认为它是相关的.

如果我做

 subprocess.call('PATH', shell=True)

然后我明白了

Traceback (most …