如何重构这个Python代码？

Question

class MainPage(webapp.RequestHandler):
  def get(self):
    user = users.get_current_user()
    tasks_query = Task.all()
    tasks = tasks_query.fetch(1000)
    if user:
      url = users.create_logout_url(self.request.uri)
    else:
      url = users.create_login_url(self.request.uri)
    template_values = {
      'tasks': tasks,
      'url': url
      }
    path = os.path.join(os.path.dirname(__file__), 'index.html')
    self.response.out.write(template.render(path, template_values))

class Gadget(webapp.RequestHandler):
  def get(self):
    user = users.get_current_user()
    tasks_query = Task.all()
    tasks = tasks_query.fetch(1000)
    if user:
      url = users.create_logout_url(self.request.uri)
    else:
      url = users.create_login_url(self.request.uri)
    template_values = {
      'tasks': tasks,
      'url': url
      }
    path = os.path.join(os.path.dirname(__file__), 'gadget.xml')
    self.response.out.write(template.render(path, template_values))

Answer 1

真的,这取决于你希望将来在两个班级之间的共同点.重构的目的是识别常见的抽象,而不是最小化代码行数.

也就是说,假设两个请求仅在模板中有所不同:

class TaskListPage(webapp.RequestHandler):
    def get(self):
        user = users.get_current_user()
        tasks_query = Task.all()
        tasks = tasks_query.fetch(1000)
        if user:
          url = users.create_logout_url(self.request.uri)
        else:
          url = users.create_login_url(self.request.uri)
        template_values = {
          'tasks': tasks,
          'url': url
          }
        path = os.path.join(os.path.dirname(__file__), self.template_name())
        self.response.out.write(template.render(path, template_values))

class MainPage(TaskListPage):
    def template_name(self):
        return 'index.html'

class Gadget(TaskListPage):
    def template_name(self):
        return 'gadget.xml'