Python 静态别名
我想在两个资源之间创建一个别名Python 静态别名,python,flask,flask-restful,Python,Flask,Flask Restful,我想在两个资源之间创建一个别名 from flask import Flask from flask_restful import Api, Resource class api_v1_help(Resource): def get(self): html_file = "API V1" return (html_file, 200, {'Content-Type': 'text/html; charset=utf-8'}) class api_v2_
from flask import Flask
from flask_restful import Api, Resource
class api_v1_help(Resource):
def get(self):
html_file = "API V1"
return (html_file, 200, {'Content-Type': 'text/html; charset=utf-8'})
class api_v2_help(Resource):
def get(self):
html_file = "API V2"
return (html_file, 200, {'Content-Type': 'text/html; charset=utf-8'})
app = Flask(__name__)
api = Api(app)
# API (current)
api.add_resource(api_v1_help, '/api/help')
# API v1
api.add_resource(api_v1_help, '/api/v1/help')
# API v2
api.add_resource(api_v2_help, '/api/v2/help')
if __name__ == '__main__':
# Start app
app.run(debug=True,port=5000)
这将导致以下错误:AssertionError:View函数映射正在覆盖现有端点函数:api_v1_帮助
我可以这样更改代码:
api.add_resource(api_v1_help, '/api/help', '/api/v1/help')
但我想知道是否有另一种使用flask restful的解决方案通过将两个API端点链接到同一个函数来处理别名
我搜索以对特定API版本的调用进行分组。请改用:
默认情况下,endpoint
设置为,因此您可以在add\u resource
调用后使用'api\u v1\u help'
作为名称。请举例说明。(复制问题的最低代码。)
# API v1
api.add_resource(api_v1_help, '/api/v1/help')
# API v2
api.add_resource(api_v2_help, '/api/v2/help')
# API (current)
app.add_url_rule('/api/help', endpoint='api_v1_help')