如何在 Django 中运行 python 脚本？

Question

我是 Django 新手，我正在尝试在脚本中导入我的模型之一，就像我们在views.py 中所做的那样。我收到错误：

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：

我的应用\呼叫中心\

CallCenter 包含__init__.py、make_call.py、models.py、views.pyMyApp 有manage.py

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

Answer 1

一般来说，最好将“临时”脚本 \xe2\x80\x93 任何可能从命令行手动运行的内容（例如 \xe2\x80\x93 ）重构为管理命令。

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这样，一旦到达您的代码，Django 运行时就会正确设置，并且您也可以免费获得命令行解析。

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你make_call.py可能会变成这样：

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呼叫中心/管理/命令/make_call.py

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from twilio.rest import Client\nfrom twilio.twiml.voice_response import VoiceResponse, Say, Dial, Number, VoiceResponse\nfrom CallCenter.models import Campaign\n\nfrom django.core.management import BaseCommand\n\n\ndef create_xml(campaign):\n    # Creates XML\n    response = VoiceResponse()\n    response.say(campaign.campaign_text)\n    return response\n\n\nclass Command(BaseCommand):\n    def add_arguments(self, parser):\n        parser.add_argument("--campaign-id", required=True, type=int)\n\n    def handle(self, campaign_id, **options):\n        campaign = Campaign.objects.get(pk=campaign_id)\n        xml = create_xml(campaign)\n        print(xml)\n

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它将被调用

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

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无论你manage.py在哪里。

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（记住和文件夹__init__.py中都有一个文件。）management/management/commands/

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