Question

我有一个django项目，正在使用Django Rest Frameowkr。我为用户模型设置了模型，序列化器，视图和URL。我有网址文件。我想在调用api URL时传递类似用户名的名称。我目前已将其设置为具有主键，因此当我输入主键时它可以工作。我想将其切换为用户名。我还希望序列化程序查询返回我传递的使用名的用户对象。我正在使用User中的Djangos标准django.contrib.auth.models对象这是我的代码

Urls.py

from django.urls import path
from django.contrib.auth.models import User

from .views import UserListView, UserDetailsView
from .views import ProfileListView, ProfileDetailsView
from .views import RoleListView, RoleDetailsView

urlpatterns = [
    path('user/', UserListView.as_view()),
    path('user/<pk>', UserDetailsView.as_view()),
]

serializer.py文件

from rest_framework import serializers

from django.contrib.auth.models import User
from users.models import Profile, Role

class UserSerializer(serializers.ModelSerializer):
    class Meta:
        model = User
        fields = ('username', 'first_name', 'last_name', 'email', 'is_staff', 'last_login')

Views.py文件

from rest_framework.generics import ListAPIView, RetrieveAPIView

from django.contrib.auth.models import User

from users.models import Profile, Role
from .serializers import UserSerializer, ProfileSerializer, RoleSerializer

class UserListView(ListAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

class UserDetailsView(RetrieveAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

Answer 1

在lookup_field中指定 UserDetailsView ，并在urls.py中更改网址格式，如下所示

# urls.py
urlpatterns = [
    path('user/', UserListView.as_view()),
    path('user/<username>', UserDetailsView.as_view()),
]


# views.py
class UserDetailsView(RetrieveAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer
    lookup_field = 'username'

尝试传递字符串以通过Django Rest Framework URL进行查询

1 个答案: