python请求-在HTTP请求中没有文件名的POST Multipart/form-data
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python requests - POST Multipart/form-data without filename in HTTP request
提问by sornars
I am trying to replicate the following POST request using the requests module in python:
我正在尝试使用 python 中的请求模块复制以下 POST 请求：
POST /example/asdfas HTTP/1.1
Host: example.com
User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:28.0) Gecko/20100101 Firefox/28.0
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-US,en;q=0.5
Accept-Encoding: gzip, deflate
Connection: keep-alive
Content-Type: multipart/form-data; boundary=---------------------------241652170216373
Content-Length: 279
-----------------------------241652170216373
Content-Disposition: form-data; name="value_1"
12345
-----------------------------241652170216373
Content-Disposition: form-data; name="value_2"
67890
-----------------------------241652170216373--
The documentation for requestssuggests that the files argument should be used.
请求的文档建议应该使用 files 参数。
When I attempt the following call:
当我尝试以下调用时：
import requests
requests.post('http://example.com/example/asdfas', files={'value_1': '12345',
'value_2': '67890'})
I get the following HTTP request:
我收到以下 HTTP 请求：
'Accept': '*/*',
'Accept-Encoding': 'gzip, deflate, compress',
'Content-Length': '264',
'User-Agent': 'python-requests/2.2.1 CPython/3.3.2 Windows/7',
'Content-Type': 'multipart/form-data; boundary=273f13699c02429db4eb95c97f757d38'
--273f13699c02429db4eb95c97f757d38
Content-Disposition: form-data; name="value_1"; filename="value_1"
12345
--273f13699c02429db4eb95c97f757d38
Content-Disposition: form-data; name="value_2"; filename="value_2"
67890
--273f13699c02429db4eb95c97f757d38--
I have also tried to use the data argument:
我也尝试使用 data 参数：
import requests
requests.post('http://example.com/example/asdfas', data={'value_1': '12345',
'value_2': '67890'})
resulting in the following HTTP request:
导致以下 HTTP 请求：
'Content-Type': 'application/x-www-form-urlencoded',
'Content-Length': '27',
'User-Agent': 'python-requests/2.2.1 CPython/3.3.2 Windows/7',
'Accept': '*/*',
'Accept-Encoding': 'gzip, deflate, compress'
value_2=67890&value_1=12345
The issue I'm having is that using the files argument results in a call that the server doesn't recognize, presumably due to the unexpected "filename" information sent in the HTTP request. Using the data argument sends the wrong Content-Type header.
我遇到的问题是，使用 files 参数会导致服务器无法识别的调用，这可能是由于 HTTP 请求中发送的意外“文件名”信息所致。使用 data 参数会发送错误的 Content-Type 标头。
The first request is known to be working on the server I wish to send the request to - what is the correct function call to identically replicate the first HTTP request?
已知第一个请求正在我希望将请求发送到的服务器上工作 - 相同地复制第一个 HTTP 请求的正确函数调用是什么？
Edit: Sample HTML form to replicate the working request:
编辑：示例 HTML 表单以复制工作请求：
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form action="http://example.com/example/asdfas" method="post" enctype="multipart/form-data">
<label for="v1">Value 1</label>
<input id="v1" type="text" name="value_1">
<label for="v2">Value 2</label>
<input id="v2" type="text" name="value_2">
<input type="submit">
</form>
</body>
</html>
采纳答案by sornars
The solution is to use tuples when passing parameters to the files argument:
解决方案是在将参数传递给 files 参数时使用元组：
import requests
requests.post('http://example.com/example/asdfas', files={'value_1': (None, '12345'), 'value_2': (None, '67890')})
Works as expected:
按预期工作：
'Accept': '*/*',
'Accept-Encoding': 'gzip, deflate, compress',
'Content-Length': '228',
'User-Agent': 'python-requests/2.2.1 CPython/3.3.2 Windows/7',
'Content-Type': 'multipart/form-data; boundary=85e90a4bbb05474ca1e23dbebdd68ed9'
--85e90a4bbb05474ca1e23dbebdd68ed9
Content-Disposition: form-data; name="value_1"
12345
--85e90a4bbb05474ca1e23dbebdd68ed9
Content-Disposition: form-data; name="value_2"
67890
--85e90a4bbb05474ca1e23dbebdd68ed9--
回答by raitisd
import requests
from requests_toolbelt import MultipartEncoder
url = 'http://example.com/example/asdfas'
fields = {'value_1':'12345', 'value_2': '67890'}
data = MultipartEncoder(fields=fields)
headers["Content-type"] = m.content_type
requests.post(url=url, data=data, headers=headers)
