我想在一个多余的config.ini中指定manage.py runserver侦听的默认端口。有没有比解析sys. exe更简单的解决方案?在manage.py和插入配置端口Argv ?

目标是运行./manage.py runserver,而不必每次都指定地址和端口,而是让它从config.ini中获取参数。


当前回答

首先你为app应用迁移

python manage.py migrate

然后:

python manage.py runserver <your port>

在浏览器运行127.0.0.1后:(你的端口)

其他回答

在上一个版本的Django(目前:4.0.3)中,你可以在settings.py文件中添加这些行


from django.core.management.commands.runserver import Command as runserver
runserver.default_port = "8000"

创建django.core.management.commands.runserver.Command的子类,并覆盖default_port成员。将文件保存为您自己的管理命令,例如在<app-name>/management/commands/runserver.py下:

from django.conf import settings
from django.core.management.commands import runserver

class Command(runserver.Command):
    default_port = settings.RUNSERVER_PORT

我在这里加载默认的端口表单设置(它反过来读取其他配置文件),但您也可以直接从其他文件读取它。

在你的项目manage.py文件中添加

from django.core.management.commands.runserver import Command as runserver

然后在def main()中:

runserver.default_port = "8001" 

我们创建了一个新的“runserver”管理命令,它是标准命令的精简包装,但改变了默认端口。粗略地说,你创建management/commands/runserver.py并放入如下内容:

# Override the value of the constant coded into django...
import django.core.management.commands.runserver as runserver
runserver.DEFAULT_PORT="8001"

# ...print out a warning...
# (This gets output twice because runserver fires up two threads (one for autoreload).
#  We're living with it for now :-)
import os
dir_path = os.path.splitext(os.path.relpath(__file__))[0]
python_path = dir_path.replace(os.sep, ".")
print "Using %s with default port %s" % (python_path, runserver.DEFAULT_PORT)

# ...and then just import its standard Command class.
# Then manage.py runserver behaves normally in all other regards.
from django.core.management.commands.runserver import Command

在.bashrc中创建环境变量 出口RUNSERVER_PORT = 8010 创建别名 alias runserver='django-admin runserver $RUNSERVER_PORT'

我使用zsh和virtualenvs包装。我把出口项目后激活脚本和分配端口为每个项目。

workon someproject
runserver