2017-09-26 195 views
1

我想让我的表单工作使用以前的计算器的答案和谷歌,但似乎没有为我工作。FOM缺少1个需要的位置参数:'请求'的形式

我有一个模型项目和一个项目团队,我希望用户能够从他创建的团队中选择一个并将其链接到项目。

我使用所谓的MYUSER

自定义的用户是我的形式,以便选择一个团队:

from django import forms 
from django.contrib.auth.models import User 
from registration.models import MyUser 
from .models import Project, Team 
from django.contrib.auth import get_user_model 

User = get_user_model() 

class EditSelectTeam(forms.Form): 

    team_choice = forms.ModelChoiceField(widget=forms.RadioSelect, queryset=None) 

    def __init__(self, User, request, *args, **kwargs): 
     super(EditSelectTeam, self).__init__(*args, **kwargs) 
     self.fields['team_choice'].queryset = Team.objects.all().filter(team_hr_admin = request.User) 

我的看法:

def TeamSelect(request): 
    if request.method == "POST": 
     select_form = EditSelectTeam(request.user, request.POST) 
     if select_form.is_valid(): 
      print('sucess') 
     else: 
      print('Fail') 

    else: 
     select_form = EditSelectTeam(request) 
    return render(request,'link_project.html', 
          {'select_form':select_form }) 

如果在我的形式,我把request.User我收到我认为的错误:

TypeError: __init__() missing 1 required positional argument: 'request' 

如果我不把用户在我__init__我得到的形式,但是当我点击文章中,我得到的错误

AttributeError: 'MyUser' object has no attribute 'user' 
+0

其他:'select_form = EditSelectTeam()'试试这个 – Akash

回答

1

__init__方法需要Userrequest

def __init__(self, User, request, *args, **kwargs): 

但只有你曾经通过其中的一个窗体:

select_form = EditSelectTeam(request.user, request.POST) 
... 
select_form = EditSelectTeam(request) 

我WOU LD改变__init__方法只取user(小写),

def __init__(self, user, *args, **kwargs): 
    super(EditSelectTeam, self).__init__(*args, **kwargs) 
    self.fields['team_choice'].queryset = Team.objects.all().filter(team_hr_admin=user) 

然后改变视图总是传递request.user

select_form = EditSelectTeam(request.user, request.POST) 
... 
select_form = EditSelectTeam(request.user) 
+0

随着编辑我得到:__ __的init()失踪1个人需要的位置参数: '请求' – Ben2pop

+0

我上面的代码不会给这个错误。 – Alasdair

+0

好了!它工作thx你;)什么完全代表要求? – Ben2pop

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