在Django中使用来自views.py的值运行python脚本

Question

我试图将我在Django的views.py文件中的值传入我写的另一个python脚本,但我不知道该怎么做.这是来自views.py文件的代码:

    def get_alarm_settings(request):
      if request.method == 'POST':
        username = request.POST.get('username')
        password = request.POST.get('password')
        alarm = Alarm.objects.all()[0].alarm_setting
        response = HttpResponse(alarm, content_type='text/plain')
        response_a = execfile('alarm_file.py')
        return response_a

我试图将响应传递给alarm_file.py,有没有人知道如何做到这一点？

Answer 1

根据问题的评论,在alarm_file.py中创建函数并导入它.要进行python alarm_file.py调用,请使用if __name__ == '__main__':alarm_file.py.它将包含调用时运行的逻辑python alarm_file.py.此外,您可以使用sys.argv获取参数,在命令行中传递.例如:

alarm_file.py:

import sys

def do_something(val):
    # do something
    print val
    # return something
    return val

if __name__ == '__main__':
    try:
        arg = sys.argv[1]
    except IndexError:
        arg = None

    return_val = do_something(arg)

这将确保您可以简单地运行: python alarm_file.py some_argument

在您的视图中只需从alarm_file导入函数并使用它:

from alarm_file import do_something
...
        response = HttpResponse(alarm, content_type='text/plain')
        response_a = do_something(response)
        return response_a