Django开发服务器可以通过编程方式启动吗? [英] Can the Django development server be started programmatically?

问题描述

我正在尝试从程序包中的另一个模块启动django开发服务器.我的模块可以导入manage.py，并且我想执行 manage.py runserver 的等效项，而不使用子进程或类似的任何东西(为什么?请参见下文).

I'm trying to start the django development server from another module in my package. My module can import manage.py, and I want to execute the equivalent of manage.py runserver without using subprocess or anything of that sort (why? see below).

目前我能想到的最好的解决方案是使用子进程:

Currently the best solution I could come up with is to use subprocess:

def run_with_default_settings():
    import inspect
    import subprocess
    currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
    subprocess.Popen(['python', 'manage.py', 'runserver'], cwd=currentdir)

但是，在我看来，此解决方案似乎过于复杂，更重要的是，它不是独立于平台的(例如，如果某人同时拥有python 2和python 3，并且定义了 python 作为python 3；或者如果在环境PATH中未定义 python 等).

However this solution seems to me rather overcomplicated, and more importantly it is not platform independent (for example if someone has both python 2 and python 3 and python is defined as python 3; or if python is not defined in the environment PATH... etc.).

我无法在线找到任何解决方案，而我尝试运行 execute_from_command_line()的每一种方法都失败了.

I couldn't find any solutions online, and every way I tried to run execute_from_command_line() failed miserably.

有什么想法吗?

推荐答案

是.只需执行 manage.py 中的内容:

Yes. Just do what's in the manage.py:

import os
from django.core.management import execute_from_command_line
os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'web.settings')
execute_from_command_line(list_of_args)

这应该可以正常工作.只需记住， execute_from_command_line 最初接受 sys.argv 作为参数，因此命令 runserver 位于索引 1 上:

This should work fine. Just remember that execute_from_command_line accepts originally sys.argv as argument, so the command runserver is on the index 1:

list_of_args = ['', 'runserver']

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