我一直在挖掘,似乎无法找到问题的答案
基本上,如果要将参数/变量传递给表单,请执行以下操作:
class SomeForm(ModelForm):
...
def __init__(self, *args, **kwargs):
my_arg = kwargs.pop('my_arg')
super(SomeForm, self).__init__(*args, **kwargs)
并在视图中传递这样的参数:
def someview(request):
form = SomeForm(request.POST, my_arg=somevalue)
if request.method == 'POST':
...
else:
form = SomeForm(my_arg=somevalue)
return render(....)
但是我正在使用inlineformset_factory
而我收到错误:
__init__() got an unexpected keyword argument 'my_arg'
如果我试图像这样传递它:
myformset = inlineformset_factory(modelA, modelB, form=SomeForm, extra=1, can_delete=True)
def someview(request):
form = myformset(request.POST, my_arg=somevalue)
if request.method == 'POST':
...
else:
form = SomeForm(my_arg=somevalue)
return render(....)
我正在使用django-dynamic-formset动态生成表单(django 1.8)
我错过了什么或者我做错了吗?如果是这样,那么什么是合理的解决方案?
更新
我试过
class BaseFormSet(BaseInlineFormSet):
def __init__(self, *args, **kwargs):
self.my_arg = kwargs.pop("my_arg")
super(BaseFormSet, self).__init__(*args, **kwargs)
然后在{/ p>中添加formset = BaseFormSet
myformset = inlineformset_factory(modelA, modelB, formset = BaseFormSet, form=SomeForm, extra=1, can_delete=True)
虽然我没有收到错误表单BaseFormSet
,但如何将my_arg
转换为SomeForm?而且我也遇到了错误:
KeyError at /someurl
'my_arg'
Request Method: GET
Request URL: http://localhost:8000/
Django Version: 1.8.2
Exception Type: KeyError
Exception Value: 'my_arg'
Exception Location: ..../forms.py in __init__, line 108(the my_arg = kwargs.pop('my_arg') line)
Python Executable: C:\Python27\python.exe
Python Version: 2.7.10
Python Path:
['C:\\Users\\lolwat\\Desktop\\ITSWEBSITE',
'C:\\Windows\\system32\\python27.zip',
'C:\\Python27\\DLLs',
'C:\\Python27\\lib',
'C:\\Python27\\lib\\plat-win',
'C:\\Python27\\lib\\lib-tk',
'C:\\Python27',
'C:\\Python27\\lib\\site-packages']
除了添加BaseFormSet
我从here
得到了这个想法