werkzeug

werkzeug: Flask框架内部本身没有实现socket，而是使用wsgi实现。wsgi是web服务网管接口，能够对请求进行封装、解析。 基于werkzeug的web应用： # 方式一：Flask返回对象是Response, 继承于werkzeug的BaseResponse。 from werkzeug.serving import run_simple from werkzeug.wrappers import BaseResponse def hello(environ, start_response): # environ, start_response参数在下面有说明 response = BaseResponse('Hello World!') return response(environ, start_response) if __name__ == '__main__': run_simple('127.0.0.1', 5000, hello) # 方式二： from werkzeug.wrappers import Request, Response @Request.application def hello(request): return Response('Hello World!') if __name__ == '__main__': from

I'm making a new website to replace a current one, using Flask micro-framework (based on Werkzeug) which uses Python (2.6 in my case). The core functionality and many pages are the same. However by using Flask many of the previous URLs are different to the old ones. I need a way to somehow store the each of the old URLs and the new URL, so that if a user types in an old URL they are simply forwarded to the new URL and everything works fine for them. Does anybody know if this is possible in Flask? Thank you in advance for your help :-) Something like this should get you started: from flask

We can obtain test_client for sample application in way like: class MyTestCase(unittest.TestCase): @classmethod def setUpClass(cls): my_app.app.config['TESTING'] = True cls.client = my_app.app.test_client() However, if we wrap app with DispatcherMiddleware - we will get error like AttributeError: 'DispatcherMiddleware' object has no attribute 'test_client' . Are there way to test composition of flask applications? We want to be able to do something like: cls.client = my_app.all_apps.test_client() When all_apps is middleware like: all_apps = DispatcherMiddleware(my_app, { '/backend': backend

I'm developping a small web service in python using: Flask (v. 0.8) storm ORM (v. 0.19) Apache with mod_wsgi I have a custom HTTP header, Unison-UUID which I'm using at some point to retrieve information in my database. here's the (slightly rewritten for simplicity) snippet that I'm having trouble with: uuid = flask.request.headers['Unison-UUID'] store = storm.locals.Store(my_database) user = store.get(models.User, uuid) The class User is more or less like this: class User(Storm): uuid = Unicode(primary=True) # Other columns.... The code above fails in the following way: File "/Users/lum

I have a Flask-based website where users can download some PDF files. This is straightforward to implement using Flask's send_file() and send_from_directory() . For example: @app.route('/downloadreport') def download_report(): return send_from_directory( '/reports', 'my_report.pdf', as_attachment=True) I'd like to perform some logic (let's call it after_download() ) AFTER the download is complete . I've tried using the @after_this_request hook. But it looks like send_file() runs asynchronously so @after_this_request may fire before the file is downloaded. For example, if the file is very large

I wanted to redirect users from test1.domain.com to test2.domain.com. I tried 'host_matching' in url_map along with 'host' in url_rule. It doesn't seem to work, shows 404 error.For example, on visiting 'localhost.com:5000' it should go to 'test.localhost.com:5000'. from flask import Flask, url_for, redirect app = Flask(__name__) app.url_map.host_matching = True @app.route("/") def hello1(): #return "Hello @ example1!" return redirect(url_for('hello2')) @app.route("/test/", host="test.localhost.com:5000") def hello2(): return "Hello @ test!" if __name__ == "__main__": app.run() Is it possible?

Flask 1, flask简介 2, flask四种主要工具 4, flask配置文件 5, flask路由 6, fbv讲解 7, fbv常用模式 8, flask之session 9, 闪现 10, 请求扩展 11, 中间件 ******************************************************************************************************************************************************************************* 1flask简介 百度百科:Flask是一个使用 Python 编写的轻量级 Web 应用框架。其 WSGI 工具箱采用 Werkzeug ， 模板引擎 则使用 Jinja2 。Flask使用 BSD 授权。 Flask是一个基于Python开发并且依赖jinja2模板和Werkzeug WSGI服务的一个微型框架，对于Werkzeug本质是Socket服务端，其用于接收http请求并对请求进行预处理，然后触发Flask框架，开发人员基于Flask框架提供的功能对请求进行相应的处理，并返回给用户，如果要返回给用户复杂的内容时，需要借助jinja2模板来实现对模板的处理，即：将模板和数据进行渲染

一:Flask简介 Flask是一个Python编写的Web 微框架，让我们可以使用Python语言快速实现一个网站或Web服务，在介绍Flask之前首先来聊下它和Django的联系以及区别，django个大而全的web框架，它内置许多模块，flask是一个小而精的轻量级框架，Django功能大而全，Flask只包含基本的配置, Django的一站式解决的思路，能让开发者不用在开发之前就在选择应用的基础设施上花费大量时间。Django有模板，表单，路由，基本的数据库管理等等内建功能。与之相反，Flask只是一个内核，默认依赖于2个外部库： Jinja2 模板引擎和 WSGI工具集--Werkzeug , flask的使用特点是基本所有的工具使用都依赖于导入的形式去扩展，flask只保留了web开发的核心功能。 WSGI（web服务器网关接口）是python中用来规定web服务器如何与python Web服务器如何与Python Web程序进行沟通的标准，本质上就是一个socket服务端。而 Werkzeug模块 就是WSGI一个具体的实现 关键词 ：一个Python编写微web框架 一个核心两个库( Jinja2 模板引擎 和 WSGI工具集) 二：Flask的优点 flask性能上基本满足一般web开发的需求, 并且灵活性以及可扩展性上要优于其他web框架,

I have some url paths and want to check if they point to a url rule in my Flask app. How can I check this using Flask? from flask import Flask, json, request, Response app = Flask('simple_app') @app.route('/foo/<bar_id>', methods=['GET']) def foo_bar_id(bar_id): if request.method == 'GET': return Response(json.dumps({'foo': bar_id}), status=200) @app.route('/bar', methods=['GET']) def bar(): if request.method == 'GET': return Response(json.dumps(['bar']), status=200) test_route_a = '/foo/1' # return foo_bar_id function test_route_b = '/bar' # return bar function davidism app.url_map stores the