如何在django中为选项标签添加属性？

Question

我必须将title属性添加到ModelChoiceField的选项中。这是我的管理员代码：

class LocModelForm(forms.ModelForm):
        def __init__(self,*args,**kwargs):
            super(LocModelForm,self).__init__(*args,**kwargs)
            self.fields['icons'] = forms.ModelChoiceField(queryset = Photo.objects.filter(galleries__title_slug = "markers"))
            self.fields['icons'].widget.attrs['class'] = 'mydds'


        class Meta:
            model = Loc
            widgets = {
                'icons' : forms.Select(attrs={'id':'mydds'}), 
                }

        class Media:
            css = {
                "all":("/media/css/dd.css",)
                }
            js=(
                '/media/js/dd.js',
                )

class LocAdmin(admin.ModelAdmin):
    form = LocModelForm

我可以为select小部件添加任何属性，但我不知道如何向选项标签添加属性。任何的想法 ？

Answer 1

首先，不要修改__init__中的字段，如果要覆盖小部件使用Meta内部类，如果要覆盖表单字段，请按正常方式声明它们（非{ -model）表格。

如果Select窗口小部件无法执行您想要的操作，那么只需创建自己的窗口小部件即可。原始窗口小部件使用render_option方法获取单个选项的HTML表示形式 - 创建子类，覆盖它，并添加您想要的任何内容。

class MySelect(forms.Select):
    def render_option(self, selected_choices, option_value, option_label):
        # look at the original for something to start with
        return u'<option whatever>...</option>'

class LocModelForm(forms.ModelForm):
    icons = forms.ModelChoiceField(
        queryset = Photo.objects.filter(galleries__title_slug = "markers"),
        widget = MySelect(attrs = {'id': 'mydds'})
    )

    class Meta:
        # ...
        # note that if you override the entire field, you don't have to override
        # the widget here
    class Media:
        # ...

Answer 2

我遇到了类似的问题，我需要动态地为每个选项添加自定义属性。但是在Django 2.0中，html渲染被移动到Widget基类中，因此修改render_option不再有效。以下是适用于我的解决方案：

from django import forms

class CustomSelect(forms.Select):
    def __init__(self, *args, **kwargs):
        self.src = kwargs.pop('src', {})
        super().__init__(*args, **kwargs)

    def create_option(self, name, value, label, selected, index, subindex=None, attrs=None):
        options = super(CustomSelect, self).create_option(name, value, label, selected, index, subindex=None, attrs=None)
        for k, v in self.src.items():
            options['attrs'][k] = v[options['value']]
        return options

class CustomForm(forms.Form):
    def __init__(self, *args, **kwargs):
        src = kwargs.pop('src', {})
        choices = kwargs.pop('choices', ())
        super().__init__(*args, **kwargs)
        if choices:
            self.fields['custom_field'].widget = CustomSelect(attrs={'class': 'some-class'}, src=src, choices=choices)

    custom_field = forms.CharField(max_length=100)

然后在视图中，使用{'form': CustomForm(choices=choices, src=src)}呈现上下文，其中src是这样的字典：{'attr-name': {'option_value': 'attr_value'}}。

Answer 3

这是我创建的一个继承自forms的类。选择（感谢Cat Plus Plus让我开始这个）。在初始化时，提供 option_title_field 参数，指示用于<option> title属性的字段。

from django import forms
from django.utils.html import escape

class SelectWithTitle(forms.Select):
    def __init__(self, attrs=None, choices=(), option_title_field=''):
        self.option_title_field = option_title_field
        super(SelectWithTitle, self).__init__(attrs, choices)

    def render_option(self, selected_choices, option_value, option_label, option_title=''):
        print option_title
        option_value = forms.util.force_unicode(option_value)
        if option_value in selected_choices:
            selected_html = u' selected="selected"'
            if not self.allow_multiple_selected:
                # Only allow for a single selection.
                selected_choices.remove(option_value)
        else:
            selected_html = ''
        return u'<option title="%s" value="%s"%s>%s</option>' % (
            escape(option_title), escape(option_value), selected_html,
            forms.util.conditional_escape(forms.util.force_unicode(option_label)))

    def render_options(self, choices, selected_choices):
            # Normalize to strings.
            selected_choices = set(forms.util.force_unicode(v) for v in selected_choices)
            choices = [(c[0], c[1], '') for c in choices]
            more_choices = [(c[0], c[1]) for c in self.choices]
            try:
                option_title_list = [val_list[0] for val_list in self.choices.queryset.values_list(self.option_title_field)]
                if len(more_choices) > len(option_title_list):
                    option_title_list = [''] + option_title_list # pad for empty label field
                more_choices = [(c[0], c[1], option_title_list[more_choices.index(c)]) for c in more_choices]
            except:
                more_choices = [(c[0], c[1], '') for c in more_choices] # couldn't get title values
            output = []
            for option_value, option_label, option_title in chain(more_choices, choices):
                if isinstance(option_label, (list, tuple)):
                    output.append(u'<optgroup label="%s">' % escape(forms.util.force_unicode(option_value)))
                    for option in option_label:
                        output.append(self.render_option(selected_choices, *option, **dict(option_title=option_title)))
                    output.append(u'</optgroup>')
                else: # option_label is just a string
                    output.append(self.render_option(selected_choices, option_value, option_label, option_title))
            return u'\n'.join(output)

class LocModelForm(forms.ModelForm):
    icons = forms.ModelChoiceField(
        queryset = Photo.objects.filter(galleries__title_slug = "markers"),
        widget = SelectWithTitle(option_title_field='FIELD_NAME_HERE')
    )

Answer 4

从django 1.11及更高版本中删除了render_option方法。看到此链接：https://docs.djangoproject.com/en/1.11/releases/1.11/#changes-due-to-the-introduction-of-template-based-widget-rendering

在这里，与Kayoz的解决方案不同。我没有像示例中那样修改名称，但我希望它仍然清楚。在模型表单中，我覆盖了该字段：

class MyForm(forms.ModelForm):
    project = ProjectModelChoiceField(label=_('Project'), widget=ProjectSelect())

然后我从上面声明类，并另外声明一个迭代器：

class ProjectModelChoiceIterator(django.forms.models.ModelChoiceIterator):
    def choice(self, obj):
        # return (self.field.prepare_value(obj), self.field.label_from_instance(obj)) #it used to be like this, but we need the extra context from the object not just the label. 
        return (self.field.prepare_value(obj), obj)

class ProjectModelChoiceField(django.forms.models.ModelChoiceField):
   def _get_choices(self):
       if hasattr(self, '_choices'):
           return self._choices
       return ProjectModelChoiceIterator(self)


class ProjectSelect(django.forms.Select):

    def create_option(self, name, value, label, selected, index, subindex=None, attrs=None):
        context = super(ProjectSelect, self).create_option(name, value, label, selected, index, subindex=None, attrs=None)

        context['attrs']['extra-attribute'] = label.extra_attribute #label is now an object, not just a string.
        return context

Answer 5

使用Django 1.11，我发现了使用已记录的API的另一种方法。如果您覆盖get_context并深入研究结构，您将在context['widget']['optgroups'][1][option_idx]['attrs']中看到各个选项属性。例如，在我的子类中，我有以下代码：

class SelectWithData(widgets.Select):
    option_data = {}

    def __init__(self, attrs=None, choices=(), option_data={}):
        super(SelectWithData, self).__init__(attrs, choices)
        self.option_data = option_data

    def get_context(self, name, value, attrs):
        context = super(SelectWithData, self).get_context(name, value, attrs)
        for optgroup in context['widget'].get('optgroups', []):
            for option in optgroup[1]:
                for k, v in six.iteritems(self.option_data.get(option['value'], {})):
                    option['attrs']['data-' + escape(k)] = escape(v)
        return context