python - 使用蓝图将参数传递给 Flask 应用程序

Question

我看过一些示例和教程，据我所知，我的观点是正确的。通过下面的基本示例，我收到以下错误:TypeError: view() got an Unexpected keywords argument 'id'

from flask import Blueprint, jsonify

location_app = Blueprint('location_app', __name__)

@location_app.route('/location/', defaults={'id': '0'})
@location_app.route('/location/<id>', methods=['GET'])
def view():
    return "location view"

Answer 1

您的函数必须采用 url 定义的变量。像这样:

@location_app.route('/location/<id>', methods=['GET'])
def view(id):
    return "location view"