Question

我是django的新手，我被要求将当前登录的用户与用户创建的Context（我的模型类）相关联。如何/在何处检索用户（来自request.user）对象？如何在模型中设置它？ 是否可以将request.user传递给使用通用视图的视图？

这是我的Context类：

from django.contrib.auth.models import User

class Context(models.Model):
    title = models.CharField(max_length=32)
    description = models.TextField()
    user = models.ForeignKey(User,null=False)

以下是表格：

class ContextForm(ModelForm):

    def __init__(self, *args, **kwargs):
        self.helper = FormHelper()
        self.helper.form_method = 'post'        
        self.helper.add_input(Submit('submit', 'Submit'))
        super(ContextForm, self).__init__(*args, **kwargs)

    class Meta:
        model = Context
        exclude = ('user')

来自urls.py的网址：

url(r'^create/$', CreateView.as_view(model=Context, form_class=ContextForm),name='context_create'),

最后，context_form.html以脆弱的形式

{% load crispy_forms_tags %}
{% block content %}
  {% crispy form %}
{% endblock %}

提前感谢您的帮助。

Answer 1

查看django-braces！它有一个form mixin从kwargs弹出当前用户。对于基于类的视图，还有mixin将用户传递给表单。如果您正在进行基于类的视图，只需使用forms.py和views.py中提供的mixins，并覆盖save（）方法。

forms.py：

from braces.forms import UserKwargModelFormMixin

class ContextForm(UserKwargModelFormMixin, ModelForm):

    def __init__(self, *args, **kwargs):
        self.helper = FormHelper()
        self.helper.form_method = 'post'        
        self.helper.add_input(Submit('submit', 'Submit'))
        super(ContextForm, self).__init__(*args, **kwargs)

    class Meta:
        model = Context
        fields = ('title', 'description',)

    def save(self, force_insert=False, force_update=False, commit=True):
        obj = super(ContextForm, self).save(commit=False)
        obj.user = self.user
        if commit:
            obj.save()
        return obj

views.py

from braces.views import UserFormKwargsMixin

class CreateContextView(UserFormKwargsMixin, CreateView)

    model = Context
    form_class = ContextForm

urls.py：

url(r'^create/$', CreateContextView.as_view(),name='context_create'),

你也可以这样做：

forms.py：

class ContextForm(ModelForm):

    def __init__(self, *args, **kwargs):
        self.user = kwargs.pop("user", None)
        self.helper = FormHelper()
        self.helper.form_method = 'post'        
        self.helper.add_input(Submit('submit', 'Submit'))
        super(ContextForm, self).__init__(*args, **kwargs)

    class Meta:
        model = Context
        fields = ('title', 'description',)

    def save(self, force_insert=False, force_update=False, commit=True):
        obj = super(ContextForm, self).save(commit=False)
        obj.user = self.user
        if commit:
            obj.save()
        return obj

views.py

def some_view(request):
    ...
    form = ContextForm(request.POST, user=request.user)
    ...

Answer 2

在你的模特中：

from django.contrib.auth.models import User

class Context(models.Model):
    title = models.CharField(max_length=32)
    description = models.TextField()
    user = models.OneToOneField(User)

在与用户关联的视图中：

    if request.method == 'POST':
            form = ContextForm(request.POST)
            if form.is_valid():
                    user = User.objects.create_user(username=form.cleaned_data['username'], email = form.cleaned_data['email'], password = form.cleaned_data['password'])
                    user.save()

                    obj = Context(user=user, title=form.cleaned_data['title'], description=form.cleaned_data['description'])
                    obj.save()

如果您需要登录，只需添加 @login_required

@login_required
def myfunction(request): 
     ...

并查看this tutorial

修改

我看过你编辑的问题，你需要将它添加到你的urls.py

TEMPLATE_CONTEXT_PROCESSORS = ( 'django.core.context_processors.request', )

这意味着它将传递给使用RequestContext呈现的所有模板，所有模板都是通用视图。

如何使用通用视图将当前登录的用户传递给模型？

2 个答案: