问题
I have a django core app called "foocore".
There are several optional pluging-like apps. For example "superfoo".
In my case every plugin adds a new choice in a model CharField which belongs to "foocore".
Django migrations detect changes if the list of choices get changed.
I think this is not necessary. At least one other developer thinks the same:
https://code.djangoproject.com/ticket/22837
class ActivePlugin(models.Model):
plugin_name = models.CharField(max_length=32, choices=get_active_plugins())
The code to get the choices:
class get_active_plugins(object):
def __iter__(self):
for item in ....:
yield item
The core "foocore" gets used in several projects and every installation has a different set of plugins. Django tries to create useless migrations ....
Is there a way to work around this?
回答1:
See this bug report and discussion for more info: https://code.djangoproject.com/ticket/22837
The proposed solution was to use a callable as the argument for choices, but it appears this has not been executed for fields but for forms only.
If you really need dynamic choices than a ForeignKey
is the best solution.
An alternative solution can be to add the requirement through a custom clean method for the field and/or creating a custom form. Form fields do support callable choices
.
See this answer for more info: https://stackoverflow.com/a/33514551/54017
回答2:
I had a similar problem with a custom field that I made for a Django 1.6 project that had the same general structure. I came to the following solution which works alright:
class ActivePluginMeta(ModelBase):
def __new__(cls, name, bases, attrs):
# Override choices attr
cls = models.base.ModelBase.__new__(cls, name, bases, attrs)
setattr(cls._meta.get_field('plugin_name'), 'choices', cls.plugin_name_choices)
return cls
class ActivePlugin(models.Model, metaclass=ActivePluginMeta):
plugin_name_choices = get_active_plugins()
plugin_name = models.CharField(max_length=32, choices=[])
That is for python 3, for python 2 you have to specify the metaclass as follows:
class ActivePlugin(models.Model):
__metaclass__ = ActivePluginMeta
plugin_name_choices = get_active_plugins()
plugin_name = models.CharField(max_length=32, choices=[])
回答3:
I had a similar problem. My choices were dynamic (all years from a starting point to the present) and every year the first time makemigrations
was run it generated new migrations for the new choice. The solution I found was customizing the field so the choices
change wouldn't be detected by makemigrations
:
from django.db import models
class YearField(models.IntegerField):
description = "A year from 2015 to the present"
def deconstruct(self):
name, path, args, kwargs = super(YearField, self).deconstruct()
# Ignore choice changes when generating migrations
kwargs.pop('choices', None)
return (name, path, args, kwargs)
来源:https://stackoverflow.com/questions/31788450/stop-django-from-creating-migrations-if-the-list-of-choices-of-a-field-changes