将烧瓶环回访问令牌

Rek*_*ekt 5 python loopback flask

我已经设置了回送API,并且我打算使用登录名,因为这样的烧瓶会发出回送请求,而回送返回一个 accessToken

例如登录到仪表板:

# Login route
@app.route("/login", methods=['GET', 'POST'])
def login():
    status = ""
    url_login = 'http://localhost:3000/api/Users/login'

    try:
        if request.method == 'POST':
            username = request.form['username']
            password =  request.form['password']


            payload_login = {
            "username": str(username),
            "password":str(password)
            }
            print(payload_login)


            r = requests.post(url_login, data=payload_login).text
            access_token = json.loads(r)

            # access_token = r['id']
            # access_token = json.loads(access_token)
            print("Access Token: " + str(access_token['id']))

            return redirect('/') #CHANGE TO 404 PAGE


    except Exception as e:
        print(e)
        return redirect('/') #CHANGE TO 404 PAGE

    return render_template('login.html')


@app.route('/dashboard', methods=['GET', 'POST'])
def logged_in_dashboard():


    return render_template('index.html')
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如何设置它,以便登录到仪表板需要环回的accessToken?在过去,我已经使用app.config['ACCESS_KEY'] ='key'并设置为如果它包含令牌,它将允许用户登录。

但是我不确定这是否是一个好习惯。您有什么建议可以处理大量用户登录的内容吗?

eku*_*ela 3

不要从 API 内部创建对 API 的请求。要在端点之间共享功能,请使用函数。这里至少需要两个函数:

  1. 函数返回有效凭证的令牌
  2. Authorization例如,需要会话或请求标头中存在令牌的函数

检查 chans 链接到的方法以获取更多实现细节:How do you Implement token authentication in Flask?

或者如何实现会话的官方教程:https ://flask.palletsprojects.com/en/1.1.x/quickstart/#sessions

其中有这样的东西:

@app.route('/')
def index():
    # this if is the login requirement
    if 'username' in session:
        return 'Logged in as %s' % escape(session['username'])
    return 'You are not logged in'

@app.route('/login', methods=['GET', 'POST'])
def login():
    if request.method == 'POST':
        # Add logic for validating username and password here.
        # If credentials are ok, set username to session.
        session['username'] = request.form['username']
        return redirect(url_for('index'))
    return '''
        <form method="post">
            <p><input type=text name=username>
            <p><input type=submit value=Login>
        </form>
    '''
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