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用于运行 Django 命令的 Python 脚本

发布于 2024-11-30 11:51:01 字数 275 浏览 0 评论 0原文

我想运行一个 python 脚本，它应该执行以下操作：

创建一个 django 项目：django-admin startproject foobar
在项目中创建一个应用程序：python manage.py barfoo
添加一个条目设置的 INSTALLED_APP 中新创建的应用 barfoo 的名称。

我怎样才能实现这个目标？

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滿滿的愛 2024-12-07 11:51:02

似乎有一种Python式的方法可以做到#1和#2

https://docs.djangoproject.com/en/dev/ref/django-admin/#running-management-commands-from-your-code

from django.core import management
management.call_command('flush', verbosity=0, interactive=False)
management.call_command('loaddata', 'test_data', verbosity=0)

There seems to be a pythonic way to do #1 and #2

https://docs.djangoproject.com/en/dev/ref/django-admin/#running-management-commands-from-your-code

from django.core import management
management.call_command('flush', verbosity=0, interactive=False)
management.call_command('loaddata', 'test_data', verbosity=0)

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人疚 2024-12-07 11:51:02

6 年后，我偶然发现了这个问题，试图弄清楚如何为应用程序编写一些测试，该应用程序仅添加与项目中其他应用程序交互的自定义模板标签。希望这可以帮助别人。

基于 @groovehunter 答案：

请注意，在调用 startapp 之前，您需要将当前目录更改为创建的项目。有关更多详细信息，请参阅此答案

from django.core import management
import os

management.call_command('startproject', 'foobar')
os.chdir('foobar')
management.call_command('startapp', 'barfoo')

，或者您可以使用 startproject 的附加参数来创建项目在当前目录下，如果确定不会有问题的话：

from django.core import management

management.call_command('startproject', 'foobar', '.')
management.call_command('startapp', 'barfoo')

6 years later I stumbled upon this question trying to figure out how to write some tests for an app which only add a custom template tag that interact with other apps in the project. Hope this can help someone.

Building on @groovehunter answer: the official documentation now (Django 1.10) inculdes this feature outside dev.

Note that you need to change current directory to the created project before call startapp. See this answer for more details

from django.core import management
import os

management.call_command('startproject', 'foobar')
os.chdir('foobar')
management.call_command('startapp', 'barfoo')

or you can use the additional argumento to startproject to create the project in the current directory, if you're sure there won't be problems:

from django.core import management

management.call_command('startproject', 'foobar', '.')
management.call_command('startapp', 'barfoo')

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蓝礼 2024-12-07 11:51:02

阅读一些有关 subprocess 和 Popen 方法的内容。这可能就是您正在寻找的。

Popen(["django-admin", "startproject", "%s" % your_name ], stdout=PIPE).communicate()
Popen(["python", "manage.py", "%s" % your_app_name ], stdout=PIPE).communicate()

3.
我知道这不是一个完美的代码，但我只是提供一个想法。

with open("settings.py", 'r') as file:
    settings = file.readlines()

new_settings = []
for line in settings:
    if "INSTALLED APPS" in line:
        new_settings.append(line.replace("INSTALLED_APPS = (", "INSTALLED_APPS = (\n'%s'," % your_app_name))
    else:
        new_settings.append(line)
with open("settings.py", 'w') as file:
    file.write("".join(new_settings))

Read a little abour subprocess and Popen method. This might be what you're looking for.

Popen(["django-admin", "startproject", "%s" % your_name ], stdout=PIPE).communicate()
Popen(["python", "manage.py", "%s" % your_app_name ], stdout=PIPE).communicate()

3.
I know that's not a perfect code, but I'm just giving an idea.

with open("settings.py", 'r') as file:
    settings = file.readlines()

new_settings = []
for line in settings:
    if "INSTALLED APPS" in line:
        new_settings.append(line.replace("INSTALLED_APPS = (", "INSTALLED_APPS = (\n'%s'," % your_app_name))
    else:
        new_settings.append(line)
with open("settings.py", 'w') as file:
    file.write("".join(new_settings))

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