一尘不染

如何使用Flask从URL获取命名参数?

flask

当用户访问在我的flask应用程序上运行的URL时,我希望Web服务能够处理问号后指定的参数:

http://10.1.1.1:5000/login?username=alex&password=pw1

#I just want to be able to manipulate the parameters
@app.route('/login', methods=['GET', 'POST'])
def login():
    username = request.form['username']
    print(username)
    password = request.form['password']
    print(password)

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2020-04-04

共1个答案

一尘不染

使用request.args得到解析查询字符串的内容:

from flask import request

@app.route(...)
def login():
    username = request.args.get('username')
    password = request.args.get('password')
2020-04-04