使用 Flask-Restplus 如何在不使 URL 与 GET 相同的情况下创建 POST api [英] With Flask-Restplus how to create a POST api without making the URL same with GET

问题描述

我是 Flask 和 Flask-RestPlus 的新手.我正在创建一个 web api，我希望我的 POST url 与 Swagger 中可见的 GET url 不同.例如在 Flask-Restplus

I'm new to Flask and Flask-RestPlus. I'm creating a web api where I want keep my POST urls different from the GET urls which are visible in Swagger. For example in Flask-Restplus

@api.route('/my_api/<int:id>')
class SavingsModeAction(Resource):
    @api.expect(MyApiModel)
    def post(self):
        pass #my code goes here

    def get(self, id):
        pass #my code goes here

因此，两个 apis url 的 swagger 看起来都像

So in swagger for the both apis url would look like

获取:/my_api/{id}

GET: /my_api/{id}

POST:/my_api/{id}

POST: /my_api/{id}

但到目前为止，我完全没有在我的 post api 中使用 {id} 部分，这可能会给用户带来一些混乱，是更新现有记录还是创建新记录，然而 api 的目的只是为了创建.

But so far I have absolutely no use of {id} part in my post api and it perhaps creates a bit of confusion for an user whether to update an existing record or to create a new, however the purpose of the api is just to create.

推荐答案

Instead of ==>获取:/my_api/{id}

使用查询参数==>获取:/my_api?id=

你上面的代码就像

Instead of ==> GET: /my_api/{id}

use query params ==> GET: /my_api?id=

your above code will be like

from flask import request

@api.route('/my_api')
class SavingsModeAction(Resource):
    
    @api.expect(MyApiModel)
    def post(self):
        id = request.args.get("id", type=int)
        ...

    def get(self):
        _id = request.args.get("_id", type=str)
        ...

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