Django形成request.user

Question

我的模型用户可以创建步枪,这种步枪显然与用户有关.

class Gun(ImageModel):
    user = models.ForeignKey(User)  
    ...  
    ...  
    ...

我有另一个依赖于此的模型并且需要使用用户步枪,但是当用户添加记录时我只想显示他的步枪.

mt模型如下所示

class Trophies(ImageModel):  
    used_his = models.ForeignKey(Gun)

我的表格如下

from django.forms import ModelForm
from django import forms
from models import Trophies
from gunsafe.models import Gun
from django.contrib.auth.models import User

class TrophiesForm(request.user, ModelForm):
    used_his = forms.ModelMultipleChoiceField(queryset=Gun.objects.filter(user__id=1))

    def __init__(self, user, *args, **kwargs):
        super(TrophiesForm, self).__init__(*args, **kwargs)
        self.fields['used_his'].queryset = User.objects.filter(pk = user)

我想知道如何获取当前登录的用户ID而不是 user__id=1

这是观点.

def edit(request, trophy_id, template_name='trophies/edit.html'):
    trophy = Trophies.objects.get(pk=trophy_id)

    if request.method == 'POST':
        form = TrophiesForm(request.POST, request.FILES, instance=trophy)
        if form.is_valid():
            newform = form.save(commit=False)
            newform.user = request.user
            newform.save()  
            ...  
            ...

Answer 1

我认为你可以通过覆盖__init__()表单的方法,传递一个实例User并过滤queryset使用该用户来实现这一点.像这样的东西:

class TrophiesForm(ModelForm):
    used_his = forms.ModelMultipleChoiceField(queryset=Gun.objects.filter(user__id=1))

    def __init__(self, user, *args, **kwargs):
        super(TrophiesForm, self).__init__(*args, **kwargs)
        self.fields['used_his'].queryset = User.objects.filter(pk = user.id)

在您的视图中,您可以传入适当的(当前登录的)实例User.

def my_trophies(request, *args, **kwargs):
    user = request.user
    form = TrophiesForm(user)
    ... 

Answer 2

Manoj顺从的另一个角度...

使用kwarg传递用户数据，以免打乱方法签名，因为它希望将request.POST作为第一个参数。更好的约定如下。

class TrophiesForm(ModelForm):
    def __init__(self, *args, **kwargs):
        #using kwargs
        user = kwargs.pop('user', None)
        super(TrophiesForm, self).__init__(*args, **kwargs)
        self.fields['used_his'].queryset = User.objects.filter(pk = user.id)

现在在通话中，这更加明确，并遵循更好的签名约定

form = TrophiesForm(request.POST, request.FILES, user=request.user)

您还可以使用instance：（在获取实例obj之前注意超级）

class TrophiesForm(ModelForm):
    def __init__(self, *args, **kwargs):
        super(SubmissionUploadForm, self).__init__(*args, **kwargs)
        user = self.instance.user
        self.fields['used_his'].queryset = User.objects.filter(pk = user.id)

然后，您可以像这样调用您的views.py：

form = TrophiesForm(instance=*MyModel*(user=request.user))