Django Todo App-按外键过滤会导致TypeError [英] Django Todo App - Filter by foreign keys results in TypeError

问题描述

这是我的模特

from django.contrib.auth.models import User

class Task(models.Model):
    owner = models.ForeignKey(User, on_delete=models.CASCADE, null=True)
    title = models.CharField(max_length=200)
    complete = models.BooleanField(default=False)
    created = models.DateTimeField(auto_now_add=True)

    def __str__(self):
        return self.title

通过以下视图和URL模式，我得到了待办事项:

With the following view and URL pattern I get my the todos:

class TaskView(viewsets.ModelViewSet):
    queryset = Task.objects.all() ## get all todos
    serializer_class = TaskSerializer

router.register("todos", views.TaskView)

但是，我想要特定所有者的待办事项.我试过了:

However, I want the todos for a specific owner. I tried this:

class TaskView(viewsets.ModelViewSet):
    queryset = Task.objects.filter(id=id)
    serializer_class = TaskSerializer

router.register("todos/<int:id>", views.TaskView)

这将导致以下错误:

TypeError: Field 'id' expected a number but got <built-in function id>.

使用 get 而不是 filter 会导致相同的问题.为什么会发生此错误，我该如何解决?

using get instead of filter resulted in the same problem. Why does this error happen and how can I solve it?

推荐答案

另一种方式:在GET parms中传递所有者ID并对其进行过滤.

Another way: pass owner id in GET parms and filter this.

serializers.py :

class TaskView(viewsets.ModelViewSet):
    serializer_class = TaskSerializer

    def get_queryset(self):
        id = self.request.GET.get('owner_id')
        queryset = Task.objects.filter(owner__id=id)
        return queryset

urls.py :

router.register("todos", views.TaskView)

在 GET 查询参数中

传递 owner_id .:

http://127.0.0.1:8000/todos/?owner_id=1

这篇关于Django Todo App-按外键过滤会导致TypeError的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！