Django ModelChoiceField optgroup 标签

Question

如何在 ModelChoiceField optgroup 标签中设置？

这是一个例子：

models.py

class Link(models.Model):
    config = models.ForeignKey(Config)
    name = models.URLField(u'Name', null=True, max_length=50)
    gateway = models.IPAddressField(u'Gateway', null=True)
    weight = models.IntegerField(u'Weight', null=True)
    description = models.TextField(u'Description', blank=True)

def __unicode__(self):
    return self.name

forms.py

class LinkForm(ModelForm):
    config = ModelChoiceField(queryset=Config.objects.all(), empty_label="Choose a link",widget=GroupedSelect())

class Meta:
    model = Link

我想像这样呈现我的 ChoiceField：

example.html

<select id="id_config" name="config">
    <option selected="selected" value="">Choose a link</option>
    <optgroup label="Configuration" >
        <option value="8">Address: 192.168.1.202/255.255.255.0 </option>
        <option value="9">Address: 192.168.1.240/255.255.255.0 </option>
        <option value="10">Address: 192.168.3.1/255.255.255.0 </option>
    </optgroup>
</select>

**更新**

我这样解决了我的问题：

class GroupedSelect(Select):
    def render(self, name, value, attrs=None, choices=()):
        if value is None: value = ''
        final_attrs = self.build_attrs(attrs, name=name)
        output = [format_html('<select{0}>', flatatt(final_attrs))]
        for index, option_gp in enumerate(self.choices):
            if index == 0:
                option_value = smart_unicode(option_gp[0])
                option_label = smart_unicode(option_gp[1])
                output.append(u'<option value="%s">%s</option>' %  (escape(option_value), escape(option_label)))
                output.append('<optgroup label = "Configuration">')
            elif index!=0 and index <= len(self.choices):
                option_value = smart_unicode(option_gp[0])
                option_label = smart_unicode(option_gp[1])
                output.append(u'<option value="%s">%s</option>' % (escape(option_value), escape(option_label)))          
        output.append(u'</optgroup>')
        output.append(u'</select>')
        return mark_safe('\n'.join(output))

Answer 1

这是一个很好的sn-p：

使用可选 Optgroups 选择字段并选择小部件： http://djangosnippets.org/snippets/200/

Answer 2

您不需要创建任何自定义字段，Django 已经完成了这项工作，只需传递格式正确的选项：

MEDIA_CHOICES = (
 ('Audio', (
   ('vinyl', 'Vinyl'),
   ('cd', 'CD'),
  )
 ),
 ('Video', (
   ('vhs', 'VHS Tape'),
   ('dvd', 'DVD'),
  )
 ),
)

Answer 3

@Stefan Manastirliu 答案的扩展，可与django-categories 一起使用。 （缺点是下面的get_tree_data() 函数只允许一级）。与bootstrap multiselect 等javascript 插件结合使用，您可以获得 等多选功能

Models.py

from categories.models import CategoryBase
class SampleCategory(CategoryBase):
    class Meta:
        verbose_name_plural = 'sample categories'

class SampleProfile(models.Model):
    categories = models.ManyToManyField('myapp.SampleCategory')

forms.py

from myapp.models import SampleCategory

    def get_tree_data():
        def rectree(toplevel):
            children_list_of_tuples = list()
            if toplevel.children.active():
                for child in toplevel.children.active():
                    children_list_of_tuples.append(tuple((child.id,child.name)))

            return children_list_of_tuples

        data = list()
        t = SampleCategory.objects.filter(active=True).filter(level=0)
        for toplevel in t:
            childrens = rectree(toplevel)
            data.append(
                tuple(
                    (
                        toplevel.name,
                        tuple(
                            childrens
                            )
                        ) 
                    )
            )
        return tuple(data)

class SampleProfileForm(forms.ModelForm):
    categories = forms.MultipleChoiceField(choices=get_tree_data())
    class Meta:
        model = SampleProfile

Answer 4

ModelChoiceField 使用ModelChoiceIterator 将查询集转换为选项列表。 您可以轻松地覆盖此类以引入组。 以下是按国家/地区对城市进行分组的示例：

from itertools import groupby
from django.forms.models import ModelChoiceField, ModelChoiceIterator
from .models import City

class CityChoiceIterator(ModelChoiceIterator):
    def __iter__(self):
        queryset = self.queryset.select_related('country').order_by('country__name', 'name')
        groups = groupby(queryset, key=lambda x: x.country)
        for country, cities in groups:
            yield [
                country.name,
                [
                    (city.id, city.name)
                    for city in cities
                ]
            ]

class CityChoiceField(ModelChoiceField):
    iterator = CityChoiceIterator

    def __init__(self, *args, **kwargs):
        super().__init__(City.objects.all(), *args, **kwargs)

注意：我没有时间检查此技术是否与 Django 3.1 中引入的新 ModelChoiceIteratorValue 兼容。