Django模型-Charfield在另一个模型中连续选择吗？

Question

我已经按照以下说明在Django中创建了一个设置表：-

class Settings(models.Model):
    name = models.CharField(max_length=200)

    class Meta:
        verbose_name = "Settings"
        verbose_name_plural = "Settings"    

    def __str__(self):
        return self.name  

class SettingChoices(models.Model):   
    setting = models.ForeignKey(Settings, on_delete=models.PROTECT) 
    choice = models.CharField(max_length=200)    
    class Meta:
        verbose_name = "Setting Choices"
        verbose_name_plural = "Setting Choices"    

    def __str__(self):
        return '{0} - {1}'.format(self.setting, self.choice)   

并将其用作示例：

Setting = Circuit Type:
Choices:
 DSL
 4G
 Fibre

然后在另一个模型中，我希望能够将此作为一组选择进行引用

class Circuits(models.Model):
    site_data = models.ForeignKey(SiteData, verbose_name="Site", on_delete=models.PROTECT)
    order_no = models.CharField(max_length=200, verbose_name="Order No")
    expected_install_date = models.DateField()
    install_date = models.DateField(blank=True, null=True)
    circuit_type = models.CharField(max_length=100, choices=*** here I would get model settings - Circuit Type - Choices ***)

目前，我在settings.py中使用了一个列表，但它并不流畅，我需要我的用户能够更改这些设置，而不是让我手动在settings.py中编辑列表并每次推送更改

我尝试了以下方法：

functions.py

def settings_circuit_types():
    from home.models import SettingChoices
    type_data = SettingChoices.objects.filter(setting__name='CIRCUIT_TYPES')
    circuit_types = []
    for c in type_data:
        circuit_types.append(c.choice)
    return circuit_types

models.py

from app.functions import settings_circuit_types
CIRCUIT_CHOICES = settings_circuit_types()
...
circuit_type = models.CharField(max_length=100, choices= CIRCUIT_CHOICES)

但这引发了错误

dango.core.exceptions.AppRegistryNotReady: Models aren't loaded yet.

这是可以理解的，我想知道通过其他方式能否实现？

谢谢

Answer 1

所以这是一种更好的方法，正如我在评论部分中提到的那样：

1-您不需要Settings和SettingChoices 。 它们基本相同，因此您可以将它们合并为一个称为Setting模型：

class Setting(models.Model):
    name = models.CharField(max_length=200)
    # if you need anything that you might think you need another model for,
    # just think this way, add a BooleanField.
    # for example if you have a setting available only for admins:

    only_admin = models.BooleanField(default=False)

    # then when you're going to make a from, just filter the options;
    # options = setting.objects.filter(only_admin=False)

    class Meta:
        verbose_name = "Settings"
        verbose_name_plural = "Settings"    

    def __str__(self):
        return self.name

2-对于Circuits模型，您只需要一个简单的ForeignKey字段：

class Circuits(models.Model):
    site_data = models.ForeignKey(SiteData, verbose_name="Site", on_delete=models.PROTECT)
    order_no = models.CharField(max_length=200, verbose_name="Order No")
    expected_install_date = models.DateField()
    install_date = models.DateField(blank=True, null=True)
    circuit_type = models.ForeignKey(Setting, null=False, blank=False)

现在，当您要为用户填写表格时：

forms.py ：

class CircuitsForm(forms.ModelForm):

    class Meta:
        model = Circuits
        fields = ('install_date', 'circuit_type') # or other fields.

    # and to filter which choices are available to choose from:
    def __init__(self, *args, **kwargs):
        super(CircuitsForm, self).__init__(*args, **kwargs)
        self.fields["circuit_type"].queryset = setting.objects.filter(only_admin=False)

这样，您就可以安全便捷地为用户和管理员创建表单。

您可以编辑管理面板本身，也可以只为管理员用户使用此类表单创建一个网址。

另外，如果您不是那种使用django呈现表单的人，则可以简单地在视图中获取可用的选项并将其传递给模板，如下所示：

settings = setting.objects.filter(only_admin=False)

并将其呈现在这样的模板中：

<select name="circuit_type">
    {% for setting in settings %}
        <option value="{{ setting.pk }}">{{ setting.name }}</option>
    {% endfor %}
</select>

现在，您只有希望它们以表单显示的选项，即使用户试图弄乱模板代码并添加更多选项，表单也不允许它们被接受，并且会为此引发错误。