Question

我有一个只有一个小部件的django表单类。我想给这个小部件命名，如<select name="custom_name"></select>，但Django获取变量的名称，并将其作为小部件的名称。例如：

class MultiSelectForm(forms.Form):
    here_is_the_name_of_the_widget = forms.MultipleChoiceField()

因此上面的代码将创建一个名为：

的小部件

<select name="here_is_the_name_of_the_widget">...</select>

因为我需要对表单进行原型设计，所以我需要在初始化时创建此名称。直到这一刻，这个表格工作正常，但标准的预先给定名称：

class MultiSelectForm(forms.Form):
    multiple_select = forms.MultipleChoiceField()

    def __init__(self, attrs=None, choices=(), *args, **kwargs):            
        self.base_fields['multiple_select'].choices = choices
        self.base_fields['multiple_select'].empty_permitted = False
        self.base_fields['multiple_select'].required = kwargs.get('required', False)
        self.base_fields['multiple_select'].widget = MultiSelect(attrs=attrs, choices=choices)

        forms.Form.__init__(self, *args, **kwargs)

所以，我想了一个方法（一个黑客）来实现我想要的但是没有像我预期的那样工作。如果我想要给小部件的名称是“countries”，则会产生以下错误：

KeyError at <project_name>
'countries'

产生该错误的代码是：

class MultiSelectForm(forms.Form):
    multiple_select = forms.MultipleChoiceField()

    def __init__(self, attrs=None, choices=(), *args, **kwargs):
        default_name = 'multiple_select'

        name = attrs.get('name', default_name)
        if name != default_name:
            setattr(self, name, forms.MultipleChoiceField())
            del attrs['name']

        '''
        The error is produced in the following line...
        '''
        self.base_fields[str(name)].choices = choices
        self.base_fields[str(name)].empty_permitted = False
        self.base_fields[str(name)].required = kwargs.get('required', False)
        self.base_fields[str(name)].widget = MultiSelect(attrs=attrs, choices=choices)

        forms.Form.__init__(self, *args, **kwargs)

有没有办法实现这一点，或更好的东西？

谢谢！

这是解决方案......

感谢@Daniel Roseman

class MultiSelectForm(forms.Form):
    def __init__(self, attrs=None, choices=(), *args, **kwargs):
        #empty_permitted is a Form's attribute
        kwargs['empty_permitted'] = False
        forms.Form.__init__(self, *args, **kwargs)

        name = attrs.get('name', 'multiple_select')
        self.fields[name] = forms.MultipleChoiceField(
            choices=choices,
            required=kwargs.get('required', False),
            widget=MultiSelect(attrs=attrs, choices=choices)
            )

Answer 1

我不确定你为什么要这么做，但你肯定是过于复杂的事情。您可以在实例化时将该字段动态添加到表单中：

class MultiSelectForm(forms.Form):
    def __init__(self, *args, **kwargs):
        dynamic_field = kwargs.pop('dynamic_field')
        super(MultiSelectForm, self).__init__(*args, **kwargs)
        self.fields[dynamic_field['name']] = forms.MultipleChoiceField(
            choices=dynamic_field['choices'], 
            empty_permitted=False
            required=dynamic_field['required'],
            widget=MultiSelect(attrs=dynamic_field['attrs']))

然后在实例化上传递字典：

my_form = MultiSelectForm(
    dynamic_field={'name': 'countries', 'choices': (mychoices), 'required': True})

Python / Django Forms：表单字段的动态变量创建和访问

这是解决方案......

1 个答案: