Question

使用requests_toolbelt以Multipart形式上传大文件，我构建了一个成功上传文件的方法，但我无法访问已发布的文件名。如何访问服务器上的文件名？

# client-side
file = open('/Volumes/Extra/test/my_video.mpg', 'rb')
payload = MultipartEncoder({file.name: file})
r = requests.post(url, data=payload, headers={'Content-Type': 'application/octet-stream'})

# server-side
@view_config(route_name='remote.agent_upload', renderer='json')
def remote_agent_upload(request):
    r = request.response
    fs = request.body_file
    f = open('/Volumes/Extra/tests2/bar.mpg', 'wb')  # wish to use filename here
    f.write(fs.read())
    fs.close()
    f.close()
    return r

Answer 1

好的，看起来您正在使用该文件的名称作为字段名称。此外，您正在这样做的方式，似乎整个帖子内容正在写入文件...这是期望的结果吗？在服务器端编写mpg文件后，您是否尝试过实际播放这些文件？

我现在没有一个HTTP服务器随时可以测试，它自动给我一个请求对象，但我假设请求对象是一个webob.Request对象（至少它看起来像是如果我错了，请纠正我。

好的，让我告诉你我的考试。（这适用于python3.4，不确定你使用的是什么版本的Python，但我认为它也适用于Python 2.7 - 虽然没有经过测试）

这个测试中的代码有点长，但是它被大量评论以帮助您理解我在每一步中所做的事情。希望它能让您更好地了解HTTP请求和响应如何使用您正在使用的工具在python中工作

# My Imports
from requests_toolbelt import MultipartEncoder
from webob import Request
import io

# Create a buffer object that can be read by the MultipartEncoder class
# This works just like an open file object
file = io.BytesIO()

# The file content will be simple for my test.
# But you could just as easily have a multi-megabyte mpg file
# Write the contents to the file
file.write(b'test mpg content')

# Then seek to the beginning of the file so that the
# MultipartEncoder can read it from the beginning
file.seek(0)

# Create the payload
payload = MultipartEncoder(
    {

        # The name of the file upload field... Not the file name
        'uploadedFile': (

            # This would be the name of the file
            'This is my file.mpg',

            # The file handle that is ready to be read from
            file,

            # The content type of the file
            'application/octet-stream'
        )
    }
)

# To send the file, you would use the requests.post method
# But the content type is not application-octet-stream
# The content type is multipart/form-data; with a boundary string
# Without the proper header type, your server would not be able to
# figure out where the file begins and ends and would think the
# entire post content is the file, which it is not. The post content
# might even contain multiple files
# So, to send your file, you would use:
#
# response = requests.post(url, data=payload, headers={'Content-Type': payload.content_type})

# Instead of sending the payload to the server,
# I am just going to grab the output as it would be sent
# This is because I don't have a server, but I can easily
# re-create the object using this output
postData = payload.to_string()

# Create an input buffer object
# This will be read by our server (our webob.Request object)
inputBuffer = io.BytesIO()

# Write the post data to the input buffer so that the webob.Request object can read it
inputBuffer.write(postData)

# And, once again, seek to 0
inputBuffer.seek(0)

# Create an error buffer so that errors can be written to it if there are any
errorBuffer = io.BytesIO()

# Setup our wsgi environment just like the server would give us
environment = {
    'HTTP_HOST': 'localhost:80',
    'PATH_INFO': '/index.py',
    'QUERY_STRING': '',
    'REQUEST_METHOD': 'POST',
    'SCRIPT_NAME': '',
    'SERVER_NAME': 'localhost',
    'SERVER_PORT': '80',
    'SERVER_PROTOCOL': 'HTTP/1.0',
    'CONTENT_TYPE': payload.content_type,
    'wsgi.errors': errorBuffer,
    'wsgi.input': inputBuffer,
    'wsgi.multiprocess': False,
    'wsgi.multithread': False,
    'wsgi.run_once': False,
    'wsgi.url_scheme': 'http',
    'wsgi.version': (1, 0)
}

# Create our request object
# This is the same as your request object and should have all our info for reading
# the file content as well as the file name
request = Request(environment)

# At this point, the request object is the same as what you get on your server
# So, from this point on, you can use the following code to get
# your actual file content as well as your file name from the object

# Our uploaded file is in the POST. And the POST field name is 'uploadedFile'
# Grab our file so that it can be read
uploadedFile = request.POST['uploadedFile']

# To read our content, you can use uploadedFile.file.read()
print(uploadedFile.file.read())

# And to get the file name, you can use uploadedFile.filename
print(uploadedFile.filename)

所以，我认为这个修改过的代码对你有用。（希望）再次，未经测试，因为我实际上没有要测试的服务器。而且，我不知道你的请求是什么类型的对象＆＃34;对象在服务器端....好的，这里是：

# client-side
import requests
file = open('/Volumes/Extra/test/my_video.mpg', 'rb')
payload = MultipartEncoder({'uploadedFile': (file.name, file, 'application/octet-stream')})
r = requests.post('http://somewhere/somefile.py', data=payload, headers={'Content-Type': payload.content_type})

# server-side
@view_config(route_name='remote.agent_upload', renderer='json')
def remote_agent_upload(request):

    # Write your actual file contents, not the post data which contains multi part boundary
    uploadedFile = request.POST['uploadedFile']
    fs = uploadedFile.file

    # The file name is insecure. What if the file name comes through as '../../../etc/passwd'
    # If you don't secure this, you've just wiped your /etc/passwd file and your server is toast
    # (assuming the web user has write permission to the /etc/passwd file
    # which it shouldn't, but just giving you a worst case scenario)
    fileName = uploadedFile.filename

    # Secure the fileName here... 
    # Make sure it doesn't have any slashes or double dots, or illegal characters, etc.
    # I'll leave that up to you

    # Write the file
    f = open('/Volumes/Extra/tests2/' + fileName, 'wb')
    f.write(fs.read())

Answer 2

原来的OP可能为时已晚，但可能会帮助其他人。这就是我使用MultipartEncoder在multipart / form-data上传中附带json的文件的方法。当我要求一个文件作为二进制文件上传时，单个json字符串作为多部分请求的一部分（因此只有两个部分，即文件和json）。请注意，在创建我的请求标头时（它是接收服务器指定的自定义标头），我从编码对象中获取content_type（它通常以multipart / form-data形式出现）。我正在使用simplejson.dumps，但我相信你可以使用json.dumps。

m = MultipartEncoder([
    ('json', (None, simplejson.dumps(datapayload), 'text/plain')),
    ('file', (os.path.basename(file_path), open(file_path, 'rb'), 'text/plain'))],
    None, encoding='utf-8')

headers = {'Authorization': 'JwToken' + ' ' + jwt_str, 'content-type': m.content_type}
response = requests.post(uri, headers=headers, data=m, timeout=45, verify=True )

在文件部分中，该字段称为“文件”，但我使用os.path.basename(file_path)从完整文件路径中获取文件名，例如c：\ temp \ mytestfile.txt。如果我愿意的话，我可以在这个字段中轻松地调用文件中的其他内容（这不是原始名称）。

Python requests_toolbelt MultipartEncoder文件名

2 个答案: