Flask 请求语法错误的问题

Question

我正在测试/试图学习烧瓶，并且正在flast_restful。我得到的此问题是：

代码400，消息不良请求语法（'name = testItem'）

main.py.py ：

from flask import Flask,request
from flask_restful import Api, Resource, reqparse

app = Flask(__name__)
api = Api(app)

product_put_args = reqparse.RequestParser()
product_put_args.add_argument("name", type = str, help = "Name of the product")
product_put_args.add_argument("quantity", type = int, help = "Quantity of the item")

products = {}

class Product(Resource):
    def get(self, barcode):
        return products[barcode]

    def put(self, barcode):
        args = product_put_args.parse_args()
        return {barcode: args}

api.add_resource(Product, "/product/<int:barcode>")

if(__name__) == "__main__":
    app.run(debug = True)

和我的 test.py

import requests

base = "http://127.0.0.1:5000/"

response = requests.put(base + "product/1", {"name": "testitem"})
print(response.json())

我试图改革两个文件，以找出要发送问题的内容，我觉得这很简单，但是如果您可以帮助我，我敢打赌，这将有所帮助我和许多其他试图开始创建REST API的人。

Answer 1

您需要默认情况下将位置信息添加到RequestParser中，它尝试从flask.request.values和flask.request.json.json中解析值，但是在您的情况下，这些值需要从a flask.request.request.form 。下面的代码解决了您的错误

from flask import Flask,request
from flask_restful import Api, Resource, reqparse

app = Flask(__name__)
api = Api(app)

product_put_args = reqparse.RequestParser()
product_put_args.add_argument("name", type = str, help = "Name of the product", location='form')
product_put_args.add_argument("quantity", type = int, help = "Quantity of the item", location='form')

products = {}

class Product(Resource):
    def get(self, barcode):
        return products[barcode]

    def put(self, barcode):
        args = product_put_args.parse_args()
        products[barcode] = args['name']
        return {barcode: args}

api.add_resource(Product, "/product/<int:barcode>")

if(__name__) == "__main__":
    app.run(debug = True)