首页 文章

django中基于类的视图中的问题SyntaxError：语法无效

提问于
浏览 次
0

我正在关注django 2.1教程，基于此链接https://docs.djangoproject.com/en/2.1/topics/class-based-views/这是我的书籍代码/ urls.py

from django.urls import path, re_path
from . import views
from book.views import BookListView

app_name = 'book'
urlpatterns = [
    path('', views.index, name = 'index')
    path('list/', BookListView.as_view(template_name="media/templates/book/book_list.html")),
]

以下是我的书/ views.py

from django.shortcuts import render

from .models import Author, Book, BookInstance, Genre

from django.views.generic import ListView

def index(request):
    num_books = Book.objects.all().count()
    num_instances = BookInstance.objects.all().count()
    num_instances_available = BookInstance.objects.filter(status__exact = 'a').count()
    num_author = Author.objects.count()

    context = {
        'num_books' : num_books,
        'num_instances' : num_instances,
        'num_instances_available' : num_instances_available,
        'num_author' : num_author,
    }
    return render(request, 'book/index.html', context) 

class BookListView(ListView):
        model = Book

我得到的错误是

文件“E：\ DJango \ mysite \ book \ urls.py”，第8行路径（'list /'，BookListView.as_view（template_name =“media / templates / book / book_list.html”）），^ SyntaxError：invalid句法

django class-based-views

2 回答

  • 2
    urlpatterns = [
    path('', views.index, name = 'index'), // missed a ,
    path('list/', BookListView.as_view(template_name="media/templates/book/book_list.html")),
]
    回复于
  • 0

    存储html文件的标准方法是在Django应用程序下创建一个名为“templates”的文件夹 . 您还可以在templates文件夹中添加另一个与app相同名称的文件夹 .

    这样你就可以像这样在你的_713509中调用这个文件：

    from django.urls import path, re_path
from . import views
from book.views import BookListView

app_name = 'book'
urlpatterns = [
    path('', views.index, name = 'index')
    path('list/', BookListView.as_view(template_name="your_app_name/book_list.html")),
]

    如果您只创建了“templates”文件夹，则可以执行以下操作：

    from django.urls import path, re_path
    from . import views
    from book.views import BookListView

    app_name = 'book'
    urlpatterns = [
        path('', views.index, name = 'index')
        path('list/', BookListView.as_view(template_name="your_app_name/book_list.html")),
    ]

    您也可以在基于类的视图中调用您正在使用的模板：

    class BookListView(ListView):
        model = Book
        template_name = 'your_app_name/book_list.html'

    或者只有“templates”文件夹：

    class BookListView(ListView):
            model = Book
            template_name = 'book_list.html'
    回复于

相关问题