使用MultipartPostHandler使用Python POST表单数据

问题:使用Python的urllib2发布数据时,所有数据都经过URL编码并作为Content-Type发送:application/x-www-form-urlencoded.上传文件时,应将Content-Type设置为multipart/form-data,并将内容编码为MIME.这个问题的讨论在这里:http: //code.activestate.com/recipes/146306/

为了解决这个限制,一些敏锐的程序员创建了一个名为MultipartPostHandler的库,它创建了一个OpenerDirector,您可以使用urllib2来主要使用multipart/form-data自动POST.此库的副本位于:http: //peerit.blogspot.com/2007/07/multipartposthandler-doesnt-work-for.html

我是Python的新手,无法让这个库工作.我基本上写了下面的代码.当我在本地HTTP代理中捕获它时,我可以看到数据仍然是URL编码的,而不是多部分MIME编码.请帮我弄清楚我做错了什么或更好的方法来完成这件事.谢谢 :-)

FROM_ADDR = 'my@email.com'

try:
    data = open(file, 'rb').read()
except:
    print "Error: could not open file %s for reading" % file
    print "Check permissions on the file or folder it resides in"
    sys.exit(1)

# Build the POST request
url = "http://somedomain.com/?action=analyze"       
post_data = {}
post_data['analysisType'] = 'file'
post_data['executable'] = data
post_data['notification'] = 'email'
post_data['email'] = FROM_ADDR

# MIME encode the POST payload
opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler)
urllib2.install_opener(opener)
request = urllib2.Request(url, post_data)
request.set_proxy('127.0.0.1:8080', 'http') # For testing with Burp Proxy

# Make the request and capture the response
try:
    response = urllib2.urlopen(request)
    print response.geturl()
except urllib2.URLError, e:
    print "File upload failed..."

编辑1:谢谢你的回复.我知道ActiveState httplib解决方案(我上面链接到它).我宁愿抽象出问题,并使用最少量的代码继续使用urllib2我是怎么做的.知道为什么没有安装和使用开瓶器？

1> Dan..：

---

似乎解决这个问题的最简单和最兼容的方法是使用'poster'模块.

# test_client.py
from poster.encode import multipart_encode
from poster.streaminghttp import register_openers
import urllib2

# Register the streaming http handlers with urllib2
register_openers()

# Start the multipart/form-data encoding of the file "DSC0001.jpg"
# "image1" is the name of the parameter, which is normally set
# via the "name" parameter of the HTML  tag.

# headers contains the necessary Content-Type and Content-Length
# datagen is a generator object that yields the encoded parameters
datagen, headers = multipart_encode({"image1": open("DSC0001.jpg")})

# Create the Request object
request = urllib2.Request("http://localhost:5000/upload_image", datagen, headers)
# Actually do the request, and get the response
print urllib2.urlopen(request).read()

这完美无缺,我不需要使用httplib.该模块可在此处获取:http: //atlee.ca/software/poster/index.html

---

@nalroff你没有调用过`poster.streaminghttp.register_openers()`

---

我知道这是一个老帖子,但是我从海报中得到了这个:`AttributeError:multipart_yielder实例没有属性'__len __'`想知道是否有其他人有这个问题.

2> nosklo..：

---

找到此配方httplib直接发布multipart (不涉及外部库)

import httplib
import mimetypes

def post_multipart(host, selector, fields, files):
    content_type, body = encode_multipart_formdata(fields, files)
    h = httplib.HTTP(host)
    h.putrequest('POST', selector)
    h.putheader('content-type', content_type)
    h.putheader('content-length', str(len(body)))
    h.endheaders()
    h.send(body)
    errcode, errmsg, headers = h.getreply()
    return h.file.read()

def encode_multipart_formdata(fields, files):
    LIMIT = '----------lImIt_of_THE_fIle_eW_$'
    CRLF = '\r\n'
    L = []
    for (key, value) in fields:
        L.append('--' + LIMIT)
        L.append('Content-Disposition: form-data; name="%s"' % key)
        L.append('')
        L.append(value)
    for (key, filename, value) in files:
        L.append('--' + LIMIT)
        L.append('Content-Disposition: form-data; name="%s"; filename="%s"' % (key, filename))
        L.append('Content-Type: %s' % get_content_type(filename))
        L.append('')
        L.append(value)
    L.append('--' + LIMIT + '--')
    L.append('')
    body = CRLF.join(L)
    content_type = 'multipart/form-data; boundary=%s' % LIMIT
    return content_type, body

def get_content_type(filename):
    return mimetypes.guess_type(filename)[0] or 'application/octet-stream'

---

这似乎是正确的,但海报模块更正确.

---

我认为这种方法更"独立",因为它不需要新模块,可以用py2exe"编译"在Windows上运行,作为*.exe文件.添加'海报'模块也可以,但不起作用.这种方法对我来说最好.欢迎来到作者!

---

文件"value"是否需要以某种方式进行编码,还是纯粹的字节流？

3> Pawel Miech..：

---

只需使用python-requests,它将设置正确的标头并为您上传:

import requests 
files = {"form_input_field_name": open("filename", "rb")}
requests.post("http://httpbin.org/post", files=files)