在django中将参数传递给动态表单 [英] passing arguments to a dynamic form in django
本文介绍了在django中将参数传递给动态表单的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这样的一些东西:
form = DynamicForm(我传递给我的表单的一些字符串参数)
这是我的形式:
class DynamicForm(Form):
def __init __(self,* args,** kwargs):
super(DynamicForm,self).__ init __(* args,** kwargs)
for range in range(5):
self.fields ['test_field_% d'%item] = CharField(max_length = 255)
解决方案
将其添加为关键字参数,称为my_arg。在调用 super()
之前,确保 pop()
关键字arg,因为父类的init方法不接受额外的关键字参数。
class DynamicForm(Form):
def __init __(self,* args,** kwargs )
my_arg = kwargs.pop('my_arg')
super(DynamicForm,self).__ init __(* args,** kwargs)
for range in range(5):
self.fields ['test_field_%d'%item] = CharField(max_length = 255)
当您创建表单时,它就像这样:
form = DynamicForm(...,my_arg ='value')
I have a Dynamic Form in forms. How can I pass an argument from my view when I instantiate my form?
Something like:
form = DynamicForm("some string argument I'm passing to my form")
This is the form I have:
class DynamicForm(Form):
def __init__(self, *args, **kwargs):
super(DynamicForm, self).__init__(*args, **kwargs)
for item in range(5):
self.fields['test_field_%d' % item] = CharField(max_length=255)
解决方案
Add it as keyword argument, say it's called my_arg. Make sure to pop()
the keyword arg before calling super()
, because the parent class's init method doesn't accept extra keyword arguments.
class DynamicForm(Form):
def __init__(self, *args, **kwargs):
my_arg = kwargs.pop('my_arg')
super(DynamicForm, self).__init__(*args, **kwargs)
for item in range(5):
self.fields['test_field_%d' % item] = CharField(max_length=255)
And when you create form it's like this:
form = DynamicForm(..., my_arg='value')
这篇关于在django中将参数传递给动态表单的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文