如何在Django中运行Python脚本？

o4hqfura 于 2023-05-16 发布在 Python

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我是Django的新手，我正试图在脚本中导入我的一个模型，就像我们在views.py中做的那样。我得到一个错误：

Traceback (most recent call last):

  File "CallCenter\make_call.py", line 3, in <module>

    from .models import Campaign

ModuleNotFoundError: No module named '__main__.models'; '__main__' is not a package

我的文件结构是这样的：
MyApp\呼叫中心
CallCenter包含__init__.py、make_call.py、models.py、views.py，MyApp包含manage.py

from twilio.rest import Client
from twilio.twiml.voice_response import VoiceResponse, Say, Dial, Number, VoiceResponse
from .models import Campaign

def create_xml():

    # Creates XML
    response = VoiceResponse()
    campaign = Campaign.objects.get(pk=1)
    response.say(campaign.campaign_text)

    return response

xml = create_xml()
print(xml)

python

来源：https://stackoverflow.com/questions/58286858/how-to-run-a-python-script-in-django

1条答案

按热度按时间

qvtsj1bj1#

一般来说，最好将“ad-hoc”脚本（例如，您可以从命令行手动运行的任何脚本）重构为management commands。
这样，一旦事情进入你的代码，Django运行时就被正确地设置了，你也可以免费获得命令行解析。
你的make_call.py可能会变成这样：

CallCenter/management/commands/make_call.py

from twilio.rest import Client
from twilio.twiml.voice_response import VoiceResponse, Say, Dial, Number, VoiceResponse
from CallCenter.models import Campaign

from django.core.management import BaseCommand

def create_xml(campaign):
    # Creates XML
    response = VoiceResponse()
    response.say(campaign.campaign_text)
    return response

class Command(BaseCommand):
    def add_arguments(self, parser):
        parser.add_argument("--campaign-id", required=True, type=int)

    def handle(self, campaign_id, **options):
        campaign = Campaign.objects.get(pk=campaign_id)
        xml = create_xml(campaign)
        print(xml)

它将被调用

$ python manage.py make_call --campaign-id=1

无论你的manage.py在哪里。
（请记住在management/和management/commands/文件夹中都有一个__init__.py文件。）

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我来回答

如何在Django中运行Python脚本？

1条答案

CallCenter/management/commands/make_call.py

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