Flask/Flask_restful 自定义 404 错误信息

Question

我有一个 flash/flask_result 应用程序设置如下。如何向整个应用程序添加自定义全局 404 错误消息？我应该用龙卷风还是烧瓶来做？

from flask import Flask
from flask_cors import CORS
from flask_restful import Api
from flask_env import MetaFlaskEnv
from tornado.httpserver import HTTPServer
from tornado.ioloop import IOLoop
from tornado.wsgi import WSGIContainer

...

#setup api app
app = Flask(__name__)
app.config.from_object(Configuration)
API = Api(app)

# allow cross site request
CORS = CORS(app, resources={r"/api/*": {"origins": "*"}})

# system endpoints
API.add_resource(Version, '/api/version')
...

if __name__ == '__main__':
    # start server
    HTTP_SERVER = HTTPServer(WSGIContainer(app))
    HTTP_SERVER.listen(port=app.config["PORT"])
    IOLoop.instance().start()

Answer 1

如果要返回自定义 404 错误页面，只需使用装饰器。

@app.errorhandler(404)
def page_not_found(e):
    return render_template('404.html'), 404