Question

因此，我正在专用服务器（Ubuntu 16.04，apache2，mod_wgsi）上运行flask应用程序。我已经在虚拟环境（python 3.5）中编译了flask_app。现在，我试图获取执行shell脚本的输出：

from flask import Flask, request
import subprocess     

app = Flask(__name__)

@app.route('/')
def index():
    test_cmd = "/venv/bin/python --version"

    try:
        result = subprocess.check_output(test_cmd, stderr=subprocess.STDOUT)

    except Exception as e:
        result = "Unexpected error: {}".format(e)
    return result

但是，我得到了错误：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/subprocess.py", line 567, in check_output
    process = Popen(stdout=PIPE, *popenargs, **kwargs)
  File "/usr/lib/python2.7/subprocess.py", line 711, in __init__
    errread, errwrite)
  File "/usr/lib/python2.7/subprocess.py", line 1343, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

apache2中默认使用哪个python？

0 个答案: