Django在选择列表更改中创建无意义的迁移 [英] Django creates pointless migrations on choices list change

问题描述

我正在尝试使用可调用的方式创建一个具有选择字段的模型，以便Django在选择列表更改时不创建迁移，如这个这个问题。

I'm trying to create a model with a choices field using a callable, so that Django doesn't create migrations when a choices list changes, as stated in this question here.

class Quote(models.Model):
    severity = models.IntegerField(choices=get_severity_choices)
    ...

class get_severity_choices(object):
    def __iter__(self):
        for item in SEVERITY_CHOICES:
            yield item

其中

SEVERITY_CHOICES = (
    (1, 'Low'),
    (2, 'Medium'),
    (3, 'High'),
)

但是，我收到一条错误消息：

However, I'm getting an error message:

quoting.Quote.severity: (fields.E004) 'choices' must be an iterable (e.g., a list or tuple).

推荐答案

我想你在混合 c c c c c c c 领域。在模型中，选项必须是可交互的 - 您不能通过callable：

I think you're mixing up the choices argument on a Model field, and that on a forms.ChoiceField field. In a model, choices must be an interable - you cannot pass a callable:

选择：一个包含
的两个项（例如[（A，B），（A，B）...]）的迭代的迭代（例如，列表或元组））用作
此字段。

choices: An iterable (e.g., a list or tuple) consisting itself of iterables of exactly two items (e.g. [(A, B), (A, B) ...]) to use as choices for this field.

您的 get_severity_choices 类不是被认为是一个迭代，因为Django希望它可以将 集合可以 ，而不仅仅是公开一个 __ iter __ 方法。

Your get_severity_choices class isn't being recognised as an iterable because Django expects it to subclass collections.Iterable rather than just expose an __iter__ method.

你可以通过 a可调用到 FormField ：

财ces ：要用作此字段的选择的二元组的迭代（例如，列表或元组），或可返回此类迭代的可调用。

choices: Either an iterable (e.g., a list or tuple) of 2-tuples to use as choices for this field, or a callable that returns such an iterable.

对于模型字段，您必须事先指定您的选择。还可以从文档中获取：

For a Model field however you must specify your choices beforehand. Also from the docs:

请注意，选择可以是任何可迭代对象必须有一个列表
或元组。这使您可以动态构建选择。但是，如果你发现
自己黑客选项是动态的，那么你可能会使用一个具有ForeignKey的正确的数据库表
。 选项适用于
静态数据，如果有任何变化，则不会发生太大变化。

Note that choices can be any iterable object – not necessarily a list or tuple. This lets you construct choices dynamically. But if you find yourself hacking choices to be dynamic, you’re probably better off using a proper database table with a ForeignKey. choices is meant for static data that doesn’t change much, if ever.

$关于为什么Django创建似乎无用的迁移，在此门票中有一些讨论：

这是设计。有几个原因，其中尤其重要的是，历史上的点数据移动需要有一个完整的
准确的模型表示，包括所有的选项，而不是
只是影响数据库的那些。

This is by design. There are several reasons, not least of which ... that datamigrations at points in history need to have a full accurate representation of the models, including all their options, not just those which affect the database.

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