自己动手写一个web框架,因为我是菜鸟,对于python的一些内建函数不是清楚,所以在写这篇文章之前需要一些python和WSGI的预备知识,这是一系列文章。这一篇只实现了如何处理url。
参考这篇文章:http://www.cnblogs.com/russellluo/p/3338616.html
预备知识
web框架主要是实现web服务器和web应用之间的交互。底层的网络协议主要有web服务器完成。譬如监听端口,填充报文等等。
Python内建函数__iter__和__call__和WSGI
迭代器iterator
迭代器为类序列对象提供了类序列的接口,也就是说类序列对象可以通过迭代器像序列一样进行迭代。说简单一点就是遍历对象。如果想让类是可迭代的,那么就必须实现__iter__和next()。
__call__
只要在类定义的时候实现了__call__方法,那么该类的对象就是可调有的,即可以将对象当做函数来使用。这里只用明白什么是__call__函数即可,因为WSGI规范中用要求。
WSGI
关于WSGI的介绍可以点击http://webpython.codepoint.net,有很详细的介绍。这里只说明一下大概。WSGI接口是用可调用的对象实现的:一个函数,一个方法或者一个可调用的实例。下面是一个实例,注释写的很详细:
# This is our application object. It could have any name, # except when using mod_wsgi where it must be "application" def application( # It accepts two arguments: # environ points to a dictionary containing CGI like environment variables # which is filled by the server for each received request from the client environ, # start_response is a callback function supplied by the server # which will be used to send the HTTP status and headers to the server start_response): # build the response body possibly using the environ dictionary response_body = 'The request method was %s' % environ['REQUEST_METHOD'] # HTTP response code and message status = '200 OK' # These are HTTP headers expected by the client. # They must be wrapped as a list of tupled pairs: # [(Header name, Header value)]. response_headers = [('Content-Type', 'text/plain'), ('Content-Length', str(len(response_body)))] # Send them to the server using the supplied function start_response(status, response_headers) # Return the response body. # Notice it is wrapped in a list although it could be any iterable. return [response_body]
简单来说就是根据接收的参数来返回相应的结果。
设计web框架
我之前用过django写过一个很简单的博客,目前放在SAE上,好久没更新了。网址:http://3.mrzysv5.sinaapp.com。一个web框架最基本的要求就是简化用户的代码量。所以在django中,我只需要写view、model和url配置文件。下面是我用django时写的一个处理视图的函数:
def blog_list(request): blogs = Article.objects.all().order_by('-publish_date') blog_num = Article.objects.count() return render_to_response('index.html', {"blogs": blogs,"blog_num":blog_num}, context_instance=RequestContext(request)) def blog_detail(request): bid = request.GET.get('id','') blog = Article.objects.get(id=bid) return render_to_response('blog.html',{'blog':blog})
需要我完成的就是操作数据库,返回相应的资源。所以我要编写的web框架就要尽可能的封装一些底层操作,留给用户一些可用的接口。根据我的观察,web框架的处理过程大致如下:
- 一个WSGI应用的基类初始化时传入配置好的url文件
- 用户写好处理方法,基类根据url调用方法
- 返回给客户端视图
一个WSGI基类,主要有以下的功能:
- 处理environ参数
- 根据url得到方法或者类名,并执行后返回
import re class WSGIapp: headers = [] def __init__(self,urls=()): self.urls = urls self.status = '200 OK' def __call__(self,environ,start_response): x = self.mapping_urls(environ) print x start_response(self.status,self.headers) if isinstance(x,str): return iter([x]) else: return iter(x) def mapping_urls(self,environ): path = environ['PATH_INFO'] for pattern,name in self.urls: m = re.match('^'+pattern+'$',path) if m: args = m.groups() func = globals()[name] return func(*args) return self.notfound() def notfound(self): self.status = '404 Not Found' self.headers = [('Content-Type','text/plain')] return '404 Not Found ' @classmethod def header(cls,name,value): cls.headers.append((name,value)) def GET_index(*args): WSGIapp.header('Content-Type','text/plain') return 'Welcome! ' def GET_hello(*args): WSGIapp.header('Content-Type','text/plain') return 'Hello %s! ' % args urls = [ ('/','GET_index'), ('/hello/(.*)','GET_hello') ] wsgiapp = WSGIapp(urls) if __name__ == '__main__': from wsgiref.simple_server import make_server httpd = make_server('',8000,wsgiapp) print 'server starting...' httpd.serve_forever()
上面的代码是不是很简介了,只需要定义函数即可。