[英]Flask-login current_user returns only ID
我設法讓燒瓶登錄工作,但是當我嘗試使用時,例如我的模板中的current_user.username
值為None
我懷疑這與我的User.get()函數有關,但在我的生活中無法理解我需要做什么,所以我可以訪問當前用戶的用戶username
或email
等等。
所以問題是,為什么我的current_user屬性的其余部分如username
,而不是填充?
注意get函數
from sqlalchemy import Column, Integer, String
from flask_login import UserMixin
from database import Base
class User(Base, UserMixin):
__tablename__ = 'users'
id = Column(Integer, primary_key=True)
username = Column(String(80), unique=True)
password = Column(String(80))
email = Column(String(120), unique=True)
def __init__(self, id):
self.id = id
def __repr__(self):
return '<User %r>' % self.username
@classmethod
def get(cls, id):
"""Return user instance of id, return None if not exist"""
try:
return cls(id)
except UserWarning:
return None
注意load_user,我相信燒瓶文檔期望它存在並且它將ID作為輸入
import flask
from flask_login import login_user, logout_user, login_required
from reportr import app, login_manager
from models import User
from forms import UploadForm, LoginForm
from database import init_session
@login_manager.user_loader
def load_user(id):
return User.get(id)
@app.route('/', methods=('GET', 'POST'))
def index():
return flask.render_template('index.html')
@app.route('/login', methods=['GET', 'POST'])
def login():
form = LoginForm()
del form.email # Don't want the email address on the form
if form.validate_on_submit():
session = init_session()
user = session.query(User).filter(User.username == flask.request.form['username']).first()
if user and user.password == flask.request.form['password']:
login_user(user)
flask.flash('Logged in successfully.', 'info')
else:
flask.flash('Username or password incorrect.', 'error')
return flask.redirect(flask.url_for('login'))
return flask.redirect(flask.url_for('index'))
return flask.render_template('login.html', form=form)
@app.route('/logout')
def logout():
logout_user()
return flask.redirect(flask.url_for('index'))
@app.route('/upload/results', methods=('GET', 'POST'))
def upload_results():
form = UploadForm()
if not form.validate_on_submit():
flask.flash('Please fill out all required fields', 'error')
return flask.render_template('upload_results.html', form=form)
@app.route("/settings")
@login_required
def settings():
pass
在這里我使用current_user.id,這是有效的。
{% block navbar %}
<nav class="navbar navbar-default" role="navigation">
<div class="container">
<div class="navbar-header">
<a class="navbar-brand" href="{{ url_for('.index') }}"><span class="glyphicon-play" aria-hidden="true"></span>Reportr!</a>
</div>
<div class="navbar-collapse collapse click-nav">
<ul class="nav navbar-nav no-js">
<li class="dropdown">
<a id='browse_menu' href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-expanded="false">Browse <span class="caret"></span></a>
<ul class="dropdown-menu" role="menu">
<li>
<a id='browse_by_project' href="#">By Project</a>
</li>
</ul>
<li class="dropdown">
<a id='upload_menu' href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-expanded="false">Upload <span class="caret"></span></a>
<ul class="dropdown-menu" role="menu">
<li>
<a id='upload_test_results' href="{{ url_for('.upload_results') }}">Test Results</a>
</li>
</ul>
</li>
</ul>
<ul class="nav navbar-nav navbar-right">
{% if current_user.is_authenticated() %}
<li class="dropdown">
<a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-expanded="false">{{ current_user.id }} <span class="caret"></span></a>
<ul class="dropdown-menu" role="menu">
<li><a href="{{ url_for('.settings') }}">Settings</a></li>
<li class="divider"></li>
<li><a href="{{ url_for('.logout') }}">Logout</a></li>
</ul>
</li>
{% else %}
<li>
<a id='login' href="{{ url_for('.login') }}">Login</a>
</li>
{% endif %}
</ul>
</div>
</div>
</nav>
{% endblock %}
我找到了問題的根源。
我的load_user()
函數沒有返回用戶,而是返回了僅填充了id
的用戶類的實例。
我更改了我的load_user函數來查詢表並返回用戶的實例,而不是類,現在它正在工作。
見下文:
@login_manager.user_loader
def load_user(user_id):
try:
return User.query.get(user_id)
except:
return None
我也擺脫了模型上的get()
函數,因為這不是必需的,而且更簡單。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.