Python实现基于HTTP文件传输实例

本文实例讲述了Python实现基于HTTP文件传输的方法。分享给大家供大家参考。具体实现方法如下：

一、问题：

因为需要最近看了一下通过POST请求传输文件的内容 并且自己写了Server和Client实现了一个简单的机遇HTTP的文件传输工具

二、实现代码：

Server端:

#coding=utf-8

from BaseHTTPServer import BaseHTTPRequestHandler

import cgi

class   PostHandler(BaseHTTPRequestHandler):

    def do_POST(self):

        form = cgi.FieldStorage(

            fp=self.rfile,

            headers=self.headers,

            environ={'REQUEST_METHOD':'POST',

                     'CONTENT_TYPE':self.headers['Content-Type'],

                     }

        )

        self.send_response(200)

        self.end_headers()

        self.wfile.write('Client: %sn ' % str(self.client_address) )

        self.wfile.write('User-agent: %sn' % str(self.headers['user-agent']))

        self.wfile.write('Path: %sn'%self.path)

        self.wfile.write('Form data:n')

        for field in form.keys():

            field_item = form[field]

            filename = field_item.filename

            filevalue  = field_item.value

            filesize = len(filevalue)#文件大小(字节)

            print len(filevalue)

            with open(filename.decode('utf-8')+'a','wb') as f:

                f.write(filevalue)

        return

if __name__=='__main__':

    from BaseHTTPServer import HTTPServer

    sever = HTTPServer(('localhost',8080),PostHandler)

    print 'Starting server, use <Ctrl-C> to stop'

    sever.serve_forever()

#coding=utf-8

import requests

url = 'http://localhost:8080'

path = u'D:快盘阿狸头像.jpg'

print path

files = {'file': open(path, 'rb')}

r = requests.post(url, files=files)

print r.url,r.text

希望本文所述对大家的Python程序设计有所帮助。