Question

我收到了错误，我不确定它是什么意思：

TypeError at / as_view() takes 1 positional argument but 2 were given

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

class Employee(models.Model):
    user = models.OneToOneField(User)
    full_name = models.CharField(max_length=50)
    phone_number = models.CharField(max_length=12)
    company = models.CharField(max_length=100)
    department = models.CharField(max_length=50)
    occupation = models.CharField(max_length=50)

    class Meta:
        verbose_name_plural = 'employees'

admin.site.register(Employee)

views.py

from django.views.generic import TemplateView
from web_app.models import Employee

class EmployeeTemplateView(TemplateView):
    model = Employee.objects.all()
    template_name = 'index.html'
    context_object_name = 'employee_view'

urls.py

from django.conf.urls import url, include
from web_app import routers
from django.contrib import admin
from . import views
from web_app.views import EmployeeTemplateView
admin.autodiscover()

router = routers.DefaultRouter()
router.register(r'employee', views.EmployeeTemplateView, "Employee")

urlpatterns = ['',
               url(r'^$', EmployeeTemplateView.as_view({'emp' : 'employees'})),
               ]

我做错了什么？

提前谢谢！

Answer 1

确保放置＆＃34; as_view（）＆＃34;因此。不是＆＃34; as_view＆＃34;。我犯了这么大的错误

Answer 2

as_view不接受位置参数，它接受关键字参数。

EmployeeTemplateView.as_view(emp='employees')

Answer 3

由于看起来您只是想将employees变量从视图传递到模板，您可以这样发送：

<强> views.py

from django.views.generic import ListView
from web_app.models import Employee

class EmployeeListView(ListView):
    model = Employee
    template_name = 'index.html'
    context_object_name = 'employees'

<强> urls.py

urlpatterns = [
               url(r'^$', EmployeeListView.as_view(), name="employees"),
               ]

然后您可以使用模板中的context_object_name：

<强>的index.html

<div>{% for employee in employees %} {{ employee }} {% endfor %}</div>

TypeError：as_view（）占用1个位置参数，但给出2个

3 个答案: