【问题标题】:How to create a formset from form-instance in Python Django如何在 Python Django 中从表单实例创建表单集
【发布时间】:2020-03-03 00:18:36
【问题描述】:

我正在尝试从表单类的实例中获取 Django - Formset。

在我的表单类中,我在 init 方法中添加了一些字段,因为表单必须提供一些灵活性。因此我不能将类作为参数传递给 formset_factory 函数。

--forms.py

class ConfigForm(forms.Form):
    def __init__(self, fields, fields_choices, *args, **kwargs):
        super(ConfigForm, self).__init__(*args, **kwargs)

        for field in fields:
            # instanciate Field from field data
            exec(
                f'self.fields["{field.name}"] ='
                f'forms.{field.field_type.field_type}('
                f'required = {field.required},'
                f'disabled = {field.disabled},'
                f'label = "{field.label}",'
                f'initial = "{field.value}",'
                f'widget = {field.widget},'
                f'help_text = "{field.description}"'
                f')'
                )
            # if field is a ChoiceField add choices to the field instance
            if 'ChoiceField' in field.field_type.field_type:
                self.fields[field.name].choices = [fields_choices[field.name]]

--views.py ...

form = forms.ConfigForm(active_fields, field_choices)
formset = formset_factory(form, extra=1)

...

但如果我尝试使用 ConfigForm 实例调用 formset_factory,则会出现以下错误:

Internal Server Error: /machines/testconfig/mw0-sap-001/
Traceback (most recent call last):
  File "C:\Users\maximilianwiederer\AppData\Local\Programs\Python\Python37\lib\site-packages\django\core\handlers\exception.py", line 34, in inner
    response = get_response(request)
  File "C:\Users\maximilianwiederer\AppData\Local\Programs\Python\Python37\lib\site-packages\django\core\handlers\base.py", line 115, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "C:\Users\maximilianwiederer\AppData\Local\Programs\Python\Python37\lib\site-packages\django\core\handlers\base.py", line 113, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\Users\maximilianwiederer\AppData\Local\Programs\Python\Python37\lib\site-packages\django\views\generic\base.py", line 71, in view
    return self.dispatch(request, *args, **kwargs)
  File "C:\Users\maximilianwiederer\AppData\Local\Programs\Python\Python37\lib\site-packages\django\views\generic\base.py", line 97, in dispatch
    return handler(request, *args, **kwargs)
  File "D:\Users\BKU\maximilianwiederer\OneDrive - Deutsche Bahn\Documents\Programmierung\Eclipse\sap4cloud-dev\machines\views.py", line 83, in get
    formset = forms.formset_factory(form, extra=1)
  File "C:\Users\maximilianwiederer\AppData\Local\Programs\Python\Python37\lib\site-packages\django\forms\formsets.py", line 441, in formset_factory
    return type(form.__name__ + 'FormSet', (formset,), attrs)
AttributeError: 'ConfigForm' object has no attribute '__name__'

【问题讨论】:

    标签: python django python-3.x django-forms formset


    【解决方案1】:

    您不能将表单的实例传递到表单集工厂。它需要一个类,而不是一个实例。但是,您可以为工厂提供一个参数字典,以提供给创建的每个表单实例。

    https://docs.djangoproject.com/en/2.2/topics/forms/formsets/#passing-custom-parameters-to-formset-forms

    以下是我建议您编写代码的方式。

    ConfigFormset = forms.formset_factory(forms.ConfigForm)
    formset = ConfigFormset(
        form_kwargs={
            'fields' : active_fields,
            'fields_choices': {}
            }
        )
    

    【讨论】:

    • 谢谢,你拯救了我的一天。我编辑了您的代码,因为您忘记了一步(实例化 ConfigFormSet,然后从中获取带有 form_kwargs 的表单集)但是现在一切正常:D
    猜你喜欢
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 2011-09-23
    • 1970-01-01
    • 1970-01-01
    • 2014-12-28
    • 2012-08-20
    • 2015-03-08
    相关资源
    最近更新 更多