龙卷风漩涡未在 URL 中拾取参数

Question

我有一个内置于 Tornado 中的 API，我正在尝试使用 tornado-swirl 对其进行记录。由于某种原因，它无法从定义的 URL 中选择可选的查询参数。这怎么能解决？我不确定我做错了什么，或者我在这里遗漏了什么。

我已经改变了模式，甚至使用了与 docs and tut.

import tornado.web
import tornado_swirl as swirl

from .base import BaseHandler

@swirl.restapi('/item/(?P<id>[\w-]+)?')
class ItemHandler(BaseHandler):
    def post(self, id):
        """Item

        Creating a new item

        Tags:
            Item
        """
        # store the item
        pass

     async def get(self, id):
        """Item

        Get items or item

        Tags:
            Item
        """
        # return all items if no id was provided
        # or return item by id when provided
        pass

我收到以下错误：

Traceback (most recent call last):
  File "/Users/.../venv/lib/python3.7/site-packages/tornado/web.py", line 1697, in _execute
    result = method(*self.path_args, **self.path_kwargs)
  File "/Users/.../venv/lib/python3.7/site-packages/tornado_swirl/views.py", line 101, in get
    for path, spec, operations in apis},
  File "/Users/.../venv/lib/python3.7/site-packages/tornado_swirl/views.py", line 100, in <dictcomp>
    'paths': {path: self.__get_api_spec(spec, operations)
  File "/Users/.../venv/lib/python3.7/site-packages/tornado_swirl/views.py", line 368, in find_api
    ['{%s}' % arg for arg in [param.name for param in vals]]
TypeError: not enough arguments for format string

显然，它没有得到参数。我认为这与我在那里定义 URL 的方式有关。

Answer 1

你只需要在文档字符串中告知路径参数，像这样：

"""Item

Creating a new item

Path Params:
    id (string) --  Your id

Tags:
    Item
"""

龙卷风漩涡未在 URL 中拾取参数

Tornado-swirl not picking up the params in a URL

python

swagger

tornado

swagger-2.0