在 migrate/makemigrations 命令的上下文中检测代码是否被 运行

Detect whether code is being run in the context of migrate/makemigrations command

我有一个带有动态选择的模型,如果我能保证在 django-admin.py migrate / makemigrations 命令以防止它创建或警告无用的选择更改。

代码:

from artist.models import Performance
from location.models import Location

def lazy_discover_foreign_id_choices():
    choices = []

    performances = Performance.objects.all()
    choices += {performance.id: str(performance) for performance in performances}.items()

    locations = Location.objects.all()
    choices += {location.id: str(location) for location in locations}.items()

    return choices
lazy_discover_foreign_id_choices = lazy(lazy_discover_foreign_id_choices, list)


class DiscoverEntry(Model):
    foreign_id = models.PositiveIntegerField('Foreign Reference', choices=lazy_discover_foreign_id_choices(), )

所以我想如果我能检测到 lazy_discover_foreign_id_choices 中的 运行 上下文,那么我可以选择输出一个空的选择列表。我正在考虑测试 sys.argv__main__.__name__,但我希望可能有更可靠的方法或 API?

我能想到的一个解决方案是将 Django makemigrations 命令子类化以在实际执行实际操作之前设置一个标志。

示例:

将该代码放入 <someapp>/management/commands/makemigrations.py,它将覆盖 Django 的默认 makemigrations 命令。

from django.core.management.commands import makemigrations
from django.db import migrations


class Command(makemigrations.Command):
    def handle(self, *args, **kwargs):
        # Set the flag.
        migrations.MIGRATION_OPERATION_IN_PROGRESS = True

        # Execute the normal behaviour.
        super(Command, self).handle(*args, **kwargs)

migrate 命令执行相同的操作。

并修改你的动态选择功能:

from django.db import migrations


def lazy_discover_foreign_id_choices():
    if getattr(migrations, 'MIGRATION_OPERATION_IN_PROGRESS', False):
        return []
    # Leave the rest as is.

它非常笨拙,但设置起来相当容易。

这是一个相当简单的方法来做到这一点(因为 django 已经为我们创建了标志):

import sys
def lazy_discover_foreign_id_choices():
    if ('makemigrations' in sys.argv or 'migrate' in sys.argv):
        return []
    # Leave the rest as is.

这应该适用于所有情况。

使用django.db.models.signals.pre_migrate应该足以检测migrate命令。缺点是你不能在配置阶段使用它。