crispy_forms.exceptions.CrispyError：| as_crispy_field传递了无效或不存在的字段

Question

我正在处理Django应用程序。在申请中存在一种缴费表格。我用django模型和django表单创建了表单。但是，当我尝试运行该应用程序时，出现以下错误。

crispy_forms.exceptions.CrispyError: |as_crispy_field got passed an invalid or inexistent field

我不确定自己在做什么错。

django models.py

CONTRIBUTE_CHOICE = (
    ('Books', 'Books'),
    ('Renovation', 'Renovation'),
    ('Other', 'Other'),
    )

class Contribute(models.Model):
    firstName = models.CharField(max_length=50)
    lastName = models.CharField(max_length=50)
    email = models.EmailField(max_length=100)
    contribution = models.CharField(max_length=100, choices=CONTRIBUTE_CHOICE)

    def publish(self):
        self.save()

    def __str__(self):
        return self.firstName

djangoforms.py

from .models import Contribute

CONTRIBUTE_CHOICE = (
    ('Books', 'Books'),
    ('Renovation', 'Renovation'),
    ('Other', 'Other'),
    )

class ContributeForm(forms.ModelForm):

    contribution = forms.ChoiceField(choices=CONTRIBUTE_CHOICE, required=True )

    class Meta:
        model = Contribute

        widgets = {
            'firstName': forms.TextInput(attrs={'placeholder': 'First Name'}),
            'lastName': forms.TextInput(attrs={'placeholder': 'Last Name'}),
            'email': forms.TextInput(attrs={'placeholder': 'Email'}),
        }
        fields = ('firstName', 'lastName', 'email', 'contribution')

django views.py

from .forms import ContributeForm

def donate(request):
    if request.method == "POST":
        contributeForm = ContributeForm(request.POST)

        if contributeForm.is_valid():
            post = contributeForm.save(commit=False)
            post.save()
            return redirect('home')

    else:
        contributeForm = ContributeForm()
        context = {'contributeForm': contributeForm}
        return render(request, 'donate.html', context)

donate.html模板

<form class='contribution_form' method="post">
    {% csrf_token %}
    <div class="row">
        <div class="col">
            {{ contributeForm.firstName|as_crispy_field }}
        </div>
        <div class="col">
            {{ contributeForm.lastName|as_crispy_field }}
        </div>
    </div>
    {{ contributeForm.email|as_crispy_field }}
    {{ contributeForm.contribution|as_crispy_field }}
    <button type="submit" class="btn btn-lg">Submit</button>
</form>

This是我创建表单所遵循的链接

[EDIT-1]

我根据答案更改了代码，现在出现以下错误。请帮助我

django.template.exceptions.TemplateSyntaxError: Invalid filter: 'as_crispy_field'

Answer 1

您没有将表单传递给donate.html模板：

def donate(request):
  if request.method == "POST":
       contributeForm = ContributeForm(request.POST)

       if contributeForm.is_valid():
          post = contributeForm.save(commit=False)
          post.save()
          return redirect('home')
       else:
       # this should be include if form validate failed
          return render(request, 'donate.html', {'contributeForm': contributeForm})

  elif request.method == "GET":
       contributeForm = ContributeForm()
       context = {'contributeForm': contributeForm}
       # return render(request, 'index.html', context) <-- why do you have this here?
       return render(request, 'donate.html', context)

Answer 2

我有同样的错误信息。这是一个ForeignKey类型的变量/表单字段，它应该显示选择，导致了问题。

Answer 3

您的表单名称是：

<块引用>

投稿表格

并在您编写的模板中：

<块引用>

投稿表格

是按键敏感的问题