Question

此问题与这些a和b类似，但差别不大。我是django的新手，所以请耐心等待。以下是我的模特

class Project(UniqueIdentifier):
    status = models.CharField(max_length=2)

class User(UniqueIdentifier):
    first_name = models.CharField(max_length=50,blank=False,null=False)
    last_name = models.CharField(max_length=50,blank=False,null=False)
    email_address = models.EmailField(max_length=200)
    user_state = models.BooleanField()

    class Meta:
        abstract = True

class Employee(User):
    experience = models.DecimalField(max_digits=4, decimal_places=2)
    userType = models.CharField(max_length=2,choices=USER_TYPES)

class Team(UniqueIdentifier):
    role = models.CharField(max_length=2)
    members = models.ManyToManyField(Employee, through="TeamMember")
    belongsTo = models.ForeignKey(Project, related_name="team")

class TeamMember(UniqueIdentifier):
    team = models.ForeignKey(Team)
    members = models.ForeignKey(Employee)

和我的通用视图类

class ProjectView(UpdateView):
    model = Project
    fset = inlineformset_factory(Project,Team,form=TeamForm,can_delete=False,fk_name="belongsTo",extra=0)
    def get_object(self, *args, **kwargs):
        c = get_object_or_404(Project, unique_id = self.kwargs.get('cid'))
        df = ProjectForm(instance=c)
        .......
        return c

    def get_context_data(self, *args, **kwargs):
        context = super(ProjectUpdateView, self).get_context_data(*args, **kwargs) 
        ..........
        return context

    def form_valid(self,form,*args, **kwargs):
        c = get_object_or_404(Project, unique_id = self.kwargs.get('cid'))
        if self.request.method == "POST":
            df = ProjectForm(self.request.POST, instance = c)
            if df.is_valid():
                c = df.save()
                fs = self.fset(self.request.POST,instance=c)
                if fs.is_valid():
                    k=fs.save()

我收到此错误。

Cannot set values on a ManyToManyField which specifies an intermediary model.

请建议是否有人有其他解决方案（我真的不想将逻辑分散到单独的表格中）。感谢您的帮助，谢谢！

**更新：找到了解决方案here。这是我的解决方案，

Class ProjectView(CreateView):
    .........
    def form_valid():
        .........
        if fs.is_valid():
                    k=fs.save(commit=False)
                    for h in k:
                        h.save()
                    for x in range(0,len(k)):
                        for y in self.request.POST.getlist('team-'+str(x)+'-members'):
                            p=TeamMember(team=k[x],members=get_object_or_404(Employee,unique_id=y))
                            p.save()

非常感谢！

Answer 1

如果你包含了一个堆栈跟踪（而不仅仅是错误）会有所帮助。但从它的外观来看，你需要保存中间模型的实例：

引自the docs：

现在您已将ManyToManyField设置为使用中间模型，您已准备好开始创建一些多对多关系。您可以通过创建中间模型的实例来完成此操作...与普通的多对多字段不同，您不能使用添加，创建或赋值来创建关系...创建此类关系的唯一方法是创建中间模型的实例。

Django Intermediary模型save（）问题

1 个答案: