无法分配“abc@gmail.com". "邀请。朋友必须是“邀请”实例

Question

我正在制定一个推荐计划。它是指一个朋友加入我们的社区。我有一个叫邀请的模型。我用了两种同型号的表格。在第一个表单中，用户必须只填写他/她的电子邮件地址。当他请求邀请时，他/她现在可以向他的/她的朋友介绍。如果被推荐的朋友点击他发送的被推荐的链接，就可以获得积分。你知道吗

我的第一个表单，即邀请函正在工作，但在请求邀请函后，如果他/她试图引用朋友填写他的电子邮件地址和他/她的朋友的电子邮件地址，我会得到一个错误

Cannot assign "abc@gmail.com". "Invitation.friend" must be a "Invitation" instance.

如何将引用的电子邮件地址（朋友）保存到数据库？你知道吗

这是我的型号.py你知道吗

class Invitation(models.Model):
    email = models.EmailField(unique=True, verbose_name=_("e-mail Address"))
    friend = models.ForeignKey('self', related_name="referral", null=True, blank=True, on_delete=models.CASCADE)
    invite_code = models.UUIDField(default=uuid.uuid4, unique=True)
    points = models.PositiveIntegerField(default=0)
    invite_accepted = models.BooleanField(verbose_name=_('invite accepted'), default=False)
    request_approved = models.BooleanField(default=True, verbose_name=_('request accepted'))

表单.py

class ReferForm(forms.Form):
    sender_email = forms.EmailField(label=_("Your email"), required=True)
    receiver_email = forms.EmailField(label=_("To email"), required=True)

    def save(self, sender_email, receiver_email):
        print('email', sender_email, receiver_email)
        new_join, created = Invitation.objects.get_or_create(email=sender_email)
        print ('new_join is', new_join, created)
        if created:
            return True
        return new_join

视图.py

class ReferInvitation(FormView):
    template_name = 'refer/refer.html'
    form_class = ReferForm

    def form_valid(self, form):
        sender_email = form.cleaned_data.get('sender_email')
        receiver_email = form.cleaned_data.get('receiver_email')
        print ('sender_email', sender_email)
        print ('friend', receiver_email)
        refer_instance = form.save(sender_email, receiver_email)
        print ('refer_instance', refer_instance)
        refer_instance.friend = receiver_email
        refer_instance.invite_code = get_invite_code()
        refer_instance.save()
        messages.success(self.request, 'You invited {0} successfully'.format(receiver_email))
        return HttpResponseRedirect('/')

一个疑问是：当他请求邀请时，发件人的电子邮件已经保存到数据库中。我的方法有效吗？你知道吗

更新

如我所说，邀请模型由两种形式使用。首先，RequestInvitation表单使用它，以便用户可以请求邀请，并且只有该用户可以使用refereform。所以在提交表单时，如果我这样做Invitation.objects.create(email=sender_email, friend=reciever_email)，我会得到

unique_constraint error on invitation_invitation.email

是的。去处理我想做的事

def save(self, sender_email, receiver_email):
    try:
        invite_instance = Invitation.objects.get(email=sender_email)
    except:
        invite_instance = None
    if invite_instance:
        invite_instance.friend = receiver_email
        return invite_instance
    return Invitation.objects.create(email=sender_email, friend=receiver_email)  

这样，如果我直接去参考表格，填写发件人和收件人的电子邮件，然后它的工作。如果我在请求邀请后转到推荐表并填写推荐表，则“朋友”字段为空。你知道吗

Answer 1

提交表单时，不要将其保存到db，将其设置为false，这样就表示您有更多的事情要处理。你知道吗

这是一个可行的解决方案，主要的变化是模型上使用的FK。你知道吗

模型。你知道吗

class Invitation(models.Model):
    email = models.EmailField(unique=True, verbose_name=("e-mail Address"))
    friend = models.EmailField(unique=True, null=True) # this was just because I didn't have any model to tie the FK to'

形式

from django import forms
from .models import Invitation


class ReferForm(forms.Form):
    sender_email = forms.EmailField(label=("Your email"), required=True)
    receiver_email = forms.EmailField(label=("To email"), required=True)

def save(self, sender_email, receiver_email):
    print('email', sender_email, receiver_email)
    #add the friend=receiver_email param as part of the argument for saving the form
    new_join, created = Invitation.objects.get_or_create(email=sender_email, friend=receiver_email)
    print ('new_join is', new_join, created)
    if created:
        return True
    return new_join

查看

from django.shortcuts import render

从.forms导入refereform 从django.http导入HttpResponseRedirect 从django.views.generic文件.编辑导入窗体视图 从.models导入邀请

class ReferInvitation(FormView):
    template_name = 'refer.html'
    form_class = ReferForm

    def form_valid(self, form):
        sender_email = form.cleaned_data.get('sender_email')
        receiver_email = form.cleaned_data.get('receiver_email')
        print ('refer_instance', refer_instance)
        #saving  and updating in the same view is not allowed in django which is why you were getting the first error
        # moved the saving of the form to the db down, you can do every other thing before saving
        form.save(sender_email, receiver_email)
        return HttpResponseRedirect('/')