Question

如何将根路径（即/）路由到视图？这是我的简单设置：

import sys
import wsgiref.simple_server

from pyramid.config import Configurator
from pyramid.response import Response
from pyramid.view import view_config

def main(argv):
    # Create Application.
    with Configurator() as config:
        app = config.make_wsgi_app()

    # Serve HTTP requests.
    server = wsgiref.simple_server.make_server('localhost', 8080, app)
    server.serve_forever()
    return 0

@view_config(name='')
def page(request):
    return Response("Root")

if __name__ == '__main__':
    sys.exit(main(sys.argv))

当我请求http://localhost:8080/时，我只是得到了404响应。从日志中：

127.0.0.1 - - [14/Apr/2018 09:14:50] "GET / HTTP/1.1" 404 153

回复机构：

<html>
 <head>
  <title>404 Not Found</title>
 </head>
 <body>
  <h1>404 Not Found</h1>
  The resource could not be found.<br/><br/>
 </body>
</html>

我正在运行Python 3.5和Pyramid 1.9.1。

Answer 1

您似乎忘记调用config.scan()，这会添加您使用@view_config注释的路线：

import sys
import wsgiref.simple_server

from pyramid.config import Configurator
from pyramid.response import Response
from pyramid.view import view_config

def main(argv):
    # Create Application.
    with Configurator() as config:
        config.scan()  # adds routes configured with the decorator
        app = config.make_wsgi_app()

    # Serve HTTP requests.
    server = wsgiref.simple_server.make_server('localhost', 8080, app)
    server.serve_forever()
    return 0

@view_config(name='')
def page(request):
    return Response("Root")


if __name__ == '__main__':
    sys.exit(main(sys.argv))

根据docs：

仅仅存在@view_config装饰器不足以执行视图配置。装饰器所做的就是使用您的配置声明“注释”该函数，它不处理它们。要使Pyramid处理pyramid.view.view_config声明，必须使用pyramid.config.Configurator的扫描方法

为什么根路径没有路由到我的视图？

1 个答案: