Django：为当前用户过滤数据库

Question

在我的网页上，我希望用户按月和年过滤数据库。

为此，我创建了一个 filters.py 。

我的问题：我没有管理用户只过滤自己的数据，但他也能看到其他用户的数据。

到目前为止，我已经尝试使用@login_required 和objects.filter(user=self.request.user) 方法。两者都没有解决我的问题。

如有任何建议，我将不胜感激！

这是我的代码：

models.py:

from django.db import models
from django.contrib.auth.models import User

class UserDetails(models.Model):
   user = models.ForeignKey(User, on_delete=models.CASCADE, related_name="new_spending", null=True) 
   expense_name = models.CharField(max_length=255)
   cost = models.FloatField()
   date_added = models.DateTimeField()

filters.py

from .models import UserDetails
import django_filters

class BudgetFilter(django_filters.FilterSet):
    year_added = django_filters.NumberFilter(field_name='date_added', lookup_expr='year', label='Year   [yyyy]')# lookup_expr='year',
    month_added = django_filters.NumberFilter(field_name='date_added', lookup_expr='month', label='Month [mm]')
    class Meta:
        model = UserDetails
        fields = ['year_added', 'month_added']

views.py

from django.shortcuts import render
from .models import UserDetails
from .filters import BudgetFilter
from django.contrib.auth.decorators import login_required

@login_required
def search(request):

    lista = UserDetails.objects.filter(user= request.user)
    filtered_list = BudgetFilter(request.GET, queryset=lista)
    return render(request, 'budget_app/user_list.html', {
        'filter': filtered_list,
        'users': lista
        })

urls.py:

from django.urls import path
from . import views 
from django_filters.views import FilterView
from .filters import BudgetFilter

urlpatterns = [
     ...
     path('search/', FilterView.as_view(filterset_class=BudgetFilter,
        template_name='budget_app/user_list.html'), name='search'),
     ]

html 文件:

{% extends 'budget_app/budget_base.html' %}



{% block content %}
  <form method="get">
    {{ filter.form.as_p }}
    <button type="submit">Search</button>
  </form>
  <ul>
  {% for item in filter.qs %}

    <li>{{ item.title }} - {{ item.date_added }}</li>
  {% endfor %}
  </ul>
{% endblock %}

谢谢！

Answer 1

你在这里触发了一个FilterView，而不是search函数，因此无论你改变什么，它都不会改变任何东西。话虽如此，我认为 subclass FilterView:

可能更好

# app/views.py

from django.contrib.auth.mixins import LoginRequiredMixin
from django_filters.views import <b>FilterView</b>
from .models import UserDetails

class UserDetailsFilterView(FilterView):
    filterset_class=BudgetFilter
    template_name='budget_app/user_list.html'

    def get_queryset(self):
        return UserDetails.objects.filter(user=self.request.user)

然后在 urls.py 你可以只使用 UserDetailsFilterView:

# app/urls.py

from django.urls import path
from . import views

urlpatterns = [
     ...
     path('search/', <b>views.UserDetailsFilterView.as_view()</b>, name='search'),
]

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