蓝图404错误处理程序不会在蓝图的URL前缀下激活

Question

我用404错误处理程序创建了一个蓝图.但是,当我在蓝图的前缀下找到不存在的URL时,会显示标准的404页面而不是我的自定义页面.如何使蓝图正确处理404错误？

以下是一个演示该问题的简短应用程序.导航到http://localhost:5000/simple/asdf不会显示蓝图的错误页面.

#!/usr/local/bin/python
# coding: utf-8

from flask import *
from config import PORT, HOST, DEBUG

simplepage = Blueprint('simple', __name__, url_prefix='/simple')

@simplepage.route('/')
def simple_root():
    return 'This simple page'

@simplepage.errorhandler(404)
def error_simple(err):
    return 'This simple error 404', err

app = Flask(__name__)
app.config.from_pyfile('config.py')
app.register_blueprint(simplepage)

@app.route('/', methods=['GET'])
def api_get():    
    return render_template('index.html')

if __name__ == '__main__':
    app.run(host=HOST,
            port=PORT,
            debug=DEBUG)

Answer 1

该文档提到404错误处理程序在蓝图上的行为不会如预期.该应用程序处理路由并在请求到达蓝图之前引发404.404处理程序仍将激活,abort(404)因为在蓝图级别进行路由后会发生这种情况.

这可以在Flask中修复(有一个关于它的公开问题).作为一种解决方法,您可以在顶级404处理程序中执行自己的错误路由.

from flask import request, render_template

@app.errorhandler(404)
def handle_404(e):
    path = request.path

    # go through each blueprint to find the prefix that matches the path
    # can't use request.blueprint since the routing didn't match anything
    for bp_name, bp in app.blueprints.items():
        if path.startswith(bp.url_prefix):
            # get the 404 handler registered by the blueprint
            handler = app.error_handler_spec.get(bp_name, {}).get(404)

            if handler is not None:
                # if a handler was found, return it's response
                return handler(e)

    # return a default response
    return render_template('404.html'), 404