Question

我正在尝试使用Popen从virtualenv执行python，但是它取自/usr/bin中的一个（至少是我想的那样，请参见下面的输出）：

In [1]: import subprocess

In [2]: proc = subprocess.Popen(['my_program', '-c', 'import sys; print sys.executable; print sys.path;'], executable='/home/user/.pypy/bin/pypy', stdout=subprocess.PIPE, stderr=subprocess.PIPE)
   ...: print proc.stdout.read(1024)
   ...: 

['', '/usr/lib/pypy/lib_pypy/__extensions__', '/usr/lib/pypy/lib_pypy', '/usr/lib/pypy/lib-python/2.7', '/usr/lib/pypy/lib-python/2.7/lib-tk', '/usr/lib/pypy/lib-python/2.7/plat-linux2', '/usr/local/lib/pypy2.7/dist-packages', '/home/user', '/usr/lib/pypy/dist-packages']


In [3]: proc = subprocess.Popen(['/home/user/.pypy/bin/pypy', '-c', 'import sys; print sys.executable; print sys.path;'], stdout=subprocess.PIPE, stderr=subprocess.PIPE)
   ...: print proc.stdout.read(1024)
   ...: 
/home/user/.pypy/bin/pypy
['', '/home/user/.pypy/lib_pypy/__extensions__', '/home/user/.pypy/lib_pypy', '/home/user/.pypy/lib-python/2.7', '/home/user/.pypy/lib-python/2.7/lib-tk', '/home/user/.pypy/lib-python/2.7/plat-linux2', '/usr/lib/pypy/lib_pypy', '/usr/lib/pypy/lib-python/2.7', '/usr/lib/pypy/lib-python/2.7/plat-linux2', '/usr/lib/pypy/lib-python/2.7/lib-tk', '/home/user/.pypy/site-packages', '/home/user']

您可以看到，没有executable参数的情况看起来都很好（可执行文件和sys.path都很好。但是，一旦我使用了参数，sys.executable为空字符串，而sys.path是系统的pypy。

我可以在没有executable参数的情况下运行，只要有另一种方法可以更改进程名称（我无法找到方法）

将可执行文件=传递给Popen时，sys.executable在子级为空

0 个答案: