Question

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命令的情况下运行代码以防止它创建或警告，我想返回一个空的选择列表无用的选择改变。

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(), )

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

Answer 1

这是一个相当非hacky的方法（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.

这适用于所有情况。

Answer 2

我能想到的解决方案是将Django makemigrations命令子类化，以便在实际执行实际操作之前设置一个标志。

示例：

将该代码放在<someapp>/management/commands/makemigrations.py中，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)

对migrate命令执行相同的操作。

修改动态选择功能：

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.

非常黑客但很容易设置。

Answer 3

使用django.db.models.signals.pre_migrate应该足以检测到<stdlib.h>命令。缺点是您不能在配置阶段使用它。

检测是否在migrate / makemigrations命令的上下文中运行代码

3 个答案: