查了众多，发现都不满意，不是这里有问题，就是那里有问题，真是郁闷，最后参照新浪微博Python SDK写了一个。花了我两个小时，毕竟初学Python，写的不好，分享一下
#/usr/bin/python#author Yee <rlk002@gmail.com>#coding=utf-8import sys,os,urllib2import mimetypesimport mimetoolsdef get_content_type(filepath): return mimetypes.guess_type(filepath)[0] or 'application/octet-stream'def encode_multipart_formdata(fields, files = {}): boundary = mimetools.choose_boundary() CRLF = '\r\n' data = [] for key in fields: data.append('--' + boundary) data.append('Content-Disposition: form-data; name="' + key + '"') data.append('') data.append(fields[key]) for key in files: data.append('--' + boundary) data.append('Content-Disposition: form-data; name="'+ key +'"; filename="'+ files[key]['filename'] + '"') data.append('Content-Type: "' + files[key]['type'] + '"') data.append('') data.append(files[key]['filedata']) data.append('--' + boundary + '--') data.append('') body = CRLF.join(data) content_type = 'multipart/form-data; boundary=%s' % boundary return {'content_type':content_type,'body':body}def http_call(url,params,files = {}): params = encode_multipart_formdata(params, files) req = urllib2.Request(url, data = params['body']) if params['content_type']: req.add_header('Content-Type',params['content_type']) resp = urllib2.urlopen(req) body = resp.read() return body def httpopenfile(url): filedata = urllib2.urlopen(url) data = filedata.read() fileinfo = filedata.info() if fileinfo.has_key("Content-Length"): filesize = fileinfo["Content-Length"] else: filesize = 0 filename = os.path.basename(url) filetype = get_content_type(filename) fileInfo = {'filename':filename,'size':filesize,'type':filetype,'filedata':data} return fileInfo def sendwb(url,data,imgpath = ''): if imgpath != '': file = httpopenfile(imgpath) ret = http_call(url,data,files = {'file':file}) return; ret = http_call(url,data)
原文地址：Python脚本POST上传数据以及文件, 感谢原作者分享。 请打开窗口，让我的灵魂与你的灵魂相拥。
