使用 MultipartPostHandler 通过 Python 发布表单数据

Question

问题：当使用 Python 的 urllib2 发布数据时，所有数据都经过 URL 编码并作为 Content-Type: application/x-www-form-urlencoded 发送。上传文件时，Content-Type 应该设置为 multipart/form-data 并且内容是 MIME 编码的。

为了解决这个限制，一些敏锐的编码人员创建了一个名为 MultipartPostHandler 的库，它创建了一个 OpenerDirector，您可以将其与 urllib2 一起使用，以自动使用 multipart/form-data 进行 POST。该库的副本在这里：MultipartPostHandler doesn't work for Unicode files

我是 Python 新手，无法让这个库正常工作。我基本上写了以下代码。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 编码的。请帮助我找出我做错了什么或更好的方法来完成这项工作。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..."

EDIT1：感谢您的回复。我知道 ActiveState httplib 解决方案（我在上面链接到它）。我宁愿抽象出问题并使用最少的代码来继续使用 urllib2。知道为什么没有安装和使用开启器吗？

Answer 1

发现这个配方可以直接使用httplib 发布多部分（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'

Answer 2

似乎解决这个问题的最简单和最兼容的方法是使用“海报”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 <input> 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()

这很完美，我不必对 httplib 感到厌烦。该模块可在此处获得： http://atlee.ca/software/poster/index.html

Answer 3

只需使用python-requests，它会设置正确的标题并为您上传：

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

Answer 4

真是巧合，我在 2 年零 6 个月前创建了这个项目

https://pypi.python.org/pypi/MultipartPostHandler2，修复了 utf-8 系统的 MultipartPostHandler。我也做了一些小改进，欢迎大家测试：）

Answer 5

我遇到了同样的问题，我需要在不使用外部库的情况下进行多部分表单发布。我写了一个完整的blogpost about the issues I ran into。

我最终使用了 http://code.activestate.com/recipes/146306/ 的修改版本。该 url 中的代码实际上只是将文件的内容附加为字符串，这可能会导致二进制文件出现问题。这是我的工作代码。

import mimetools
import mimetypes
import io
import http
import json


form = MultiPartForm()
form.add_field("form_field", "my awesome data")

# Add a fake file     
form.add_file(key, os.path.basename(filepath),
    fileHandle=codecs.open("/path/to/my/file.zip", "rb"))

# Build the request
url = "http://www.example.com/endpoint"
schema, netloc, url, params, query, fragments = urlparse.urlparse(url)

try:
    form_buffer =  form.get_binary().getvalue()
    http = httplib.HTTPConnection(netloc)
    http.connect()
    http.putrequest("POST", url)
    http.putheader('Content-type',form.get_content_type())
    http.putheader('Content-length', str(len(form_buffer)))
    http.endheaders()
    http.send(form_buffer)
except socket.error, e:
    raise SystemExit(1)

r = http.getresponse()
if r.status == 200:
    return json.loads(r.read())
else:
    print('Upload failed (%s): %s' % (r.status, r.reason))

class MultiPartForm(object):
    """Accumulate the data to be used when posting a form."""

    def __init__(self):
        self.form_fields = []
        self.files = []
        self.boundary = mimetools.choose_boundary()
        return

    def get_content_type(self):
        return 'multipart/form-data; boundary=%s' % self.boundary

    def add_field(self, name, value):
        """Add a simple field to the form data."""
        self.form_fields.append((name, value))
        return

    def add_file(self, fieldname, filename, fileHandle, mimetype=None):
        """Add a file to be uploaded."""
        body = fileHandle.read()
        if mimetype is None:
            mimetype = mimetypes.guess_type(filename)[0] or 'application/octet-stream'
        self.files.append((fieldname, filename, mimetype, body))
        return

    def get_binary(self):
        """Return a binary buffer containing the form data, including attached files."""
        part_boundary = '--' + self.boundary

        binary = io.BytesIO()
        needsCLRF = False
        # Add the form fields
        for name, value in self.form_fields:
            if needsCLRF:
                binary.write('\r\n')
            needsCLRF = True

            block = [part_boundary,
              'Content-Disposition: form-data; name="%s"' % name,
              '',
              value
            ]
            binary.write('\r\n'.join(block))

        # Add the files to upload
        for field_name, filename, content_type, body in self.files:
            if needsCLRF:
                binary.write('\r\n')
            needsCLRF = True

            block = [part_boundary,
              str('Content-Disposition: file; name="%s"; filename="%s"' % \
              (field_name, filename)),
              'Content-Type: %s' % content_type,
              ''
              ]
            binary.write('\r\n'.join(block))
            binary.write('\r\n')
            binary.write(body)


        # add closing boundary marker,
        binary.write('\r\n--' + self.boundary + '--\r\n')
        return binary

Answer 6

为了回答 OP 关于为什么原始代码不起作用的问题，传入的处理程序不是类的实例。线

# MIME encode the POST payload
opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler)

应该阅读

opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler())