python - 我如何在 Pyramid 中建立一条路线

Question

我使用pyramid_handlers进行路由，controller.py:

import pyramid_handlers
from blue_yellow_app.controllers.base_controller import BaseController
from blue_yellow_app.services.albums_service import AlbumsService


class AlbumsController(BaseController):
    @pyramid_handlers.action(renderer='templates/albums/index.pt')
    def index(self):
        # data / service access
        all_albums = AlbumsService.get_albums()

        # return the model
        return {'albums': all_albums}

我已经在 __init__.py 中注册了，如下所示:

from pyramid.config import Configurator
import blue_yellow_app.controllers.controller as albums


def main(_, **settings):
    config = Configurator(settings=settings)
    config.include('pyramid_chameleon')
    config.include('pyramid_handlers')

    config.add_handler(
        'albums' + 'ctrl_index', '/' + 'albums',
        handler=albums.AlbumsController, action='index')
    config.add_handler(
        'albums' + 'ctrl_index/', '/' + 'albums' + '/',
        handler=albums.AlbumsController, action='index')
    config.add_handler(
        'albums' + 'ctrl', '/' + 'albums' + '/{id}',
        handler=albums.AlbumsController)

现在如何为某个专辑添加新的 Controller View ？我试过了添加新 View ，如下所示:

import pyramid_handlers
from blue_yellow_app.controllers.base_controller import BaseController
from blue_yellow_app.services.albums_service import AlbumsService


class AlbumsController(BaseController):
    @pyramid_handlers.action(renderer='templates/albums/index.pt')
    def index(self):
        # data / service access
        all_albums = AlbumsService.get_albums()

        # return the model
        return {'albums': all_albums}

    @pyramid_handlers.action(
        name='albums/{id}',
        renderer='templates/albums/item.pt')
    def album(self):
        print ('test')
        return {}

但是这不起作用。如何添加路由albums/{id}的 View ？

Answer 1

看起来这段代码来 self 的 Python for Entrepreneurs course 。让我们关注 add_handler 部分。该函数的通用形式为:

 config.add_handler(NAME, URL, handler=HANDLER, action=OPTIONAL_ACTION_METHOD)

您想要将 URL /albums/rock-classics 映射到操作方法 def album(self)。在 add_handler 调用中，您有:

config.add_handler(
    'albumsctrl', '/' + 'albums' + '/{id}',
    handler=albums.AlbumsController)

问题有两个:

您没有在路由值或函数调用中指定操作。你应该有:

# via add_handler, url: /albums/rock-classics
config.add_handler(
     'albumsctrl', '/albums/{id}',
     handler=albums.AlbumsController, action=album)

或

# via route, url: /albums/album/rock-classics
config.add_handler(
     'albumsctrl', '/albums/{action}/{id}',
     handler=albums.AlbumsController)

第二个问题是操作方法本身中的您的名字

 @pyramid_handlers.action(
     name='albums/{id}', <----- PROBLEM: this is not a valid action name
     renderer='templates/albums/item.pt')
 def album(self):
     print ('test')
     return {}

它应该是重复的 name='album' 或只是方法的名称:

 @pyramid_handlers.action(renderer='templates/albums/item.pt')
 def album(self):
     print ('test')
     return {}