这是在多部分请求中发送文件的一种方法

import requests
headers = {"Authorization": "Bearer <token>"}
myfile = 'file.txt'
myfile2 = {'file': (myfile, open(myfile, 'rb'),'application/octet-stream')}
url = 'https://example.com/path'
r = requests.post(url, files=myfile2, headers=headers,verify=False)
print(r.content)

其他方法

import requests

url = "https://example.com/path"

payload={}
files=[
  ('file',('file',open('/path/to/file','rb'),'application/octet-stream'))
]
headers = {
  'Authorization': 'Bearer <token>'
}

response = requests.request("POST", url, headers=headers, data=payload, files=files)

print(response.text)

我都测试过了，两者都很好。

邮差生成的代码文件上传附加的表单字段:

import http.client
import mimetypes
from codecs import encode

conn = http.client.HTTPSConnection("data.XXXX.com")
dataList = []
boundary = 'wL36Yn8afVp8Ag7AmP8qZ0SA4n1v9T'
dataList.append(encode('--' + boundary))
dataList.append(encode('Content-Disposition: form-data; name=batchSize;'))

dataList.append(encode('Content-Type: {}'.format('text/plain')))
dataList.append(encode(''))

dataList.append(encode("1"))
dataList.append(encode('--' + boundary))
dataList.append(encode('Content-Disposition: form-data; name=file; filename={0}'.format('FileName-1.json')))

fileType = mimetypes.guess_type('FileName-1.json')[0] or 'application/octet-stream'
dataList.append(encode('Content-Type: {}'.format(fileType)))
dataList.append(encode(''))

with open('FileName-1.json', 'rb') as f:
  dataList.append(f.read())
dataList.append(encode('--'+boundary+'--'))
dataList.append(encode(''))
body = b'\r\n'.join(dataList)
payload = body
headers = {
  'Cookie': 'XXXXXXXXXXX',
  'Content-type': 'multipart/form-data; boundary={}'.format(boundary)
}
conn.request("POST", "/fileupload/uri/XXXX", payload, headers)
res = conn.getresponse()
data = res.read()
print(data.decode("utf-8"))

我试图用python 3中的请求模块发送一个请求到URL_server。 这对我来说很管用:

# -*- coding: utf-8 *-*
import json, requests

URL_SERVER_TO_POST_DATA = "URL_to_send_POST_request"
HEADERS = {"Content-Type" : "multipart/form-data;"}

def getPointsCC_Function():
  file_data = {
      'var1': (None, "valueOfYourVariable_1"),
      'var2': (None, "valueOfYourVariable_2")
  }

  try:
    resElastic = requests.post(URL_GET_BALANCE, files=file_data)
    res = resElastic.json()
  except Exception as e:
    print(e)

  print (json.dumps(res, indent=4, sort_keys=True))

getPointsCC_Function()

地点:

URL_SERVER_TO_POST_DATA =我们要发送数据的服务器 报头=发送的报头 file_data =发送的参数

发送multipart/form-data键和值

curl命令:

curl -X PUT http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F taskStatus=1

python请求-更复杂的POST请求:

    updateTaskUrl = "http://127.0.0.1:8080/api/xxx"
    updateInfoDict = {
        "taskStatus": 1,
    }
    resp = requests.put(updateTaskUrl, data=updateInfoDict)

发送多部分/表单数据文件

curl命令:

curl -X POST http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F file=@/Users/xxx.txt

POST一个多部分编码的文件:

    filePath = "/Users/xxx.txt"
    fileFp = open(filePath, 'rb')
    fileInfoDict = {
        "file": fileFp,
    }
    resp = requests.post(uploadResultUrl, files=fileInfoDict)

这是所有。

下面是使用请求上传带有附加参数的单个文件的简单代码片段:

url = 'https://<file_upload_url>'
fp = '/Users/jainik/Desktop/data.csv'

files = {'file': open(fp, 'rb')}
payload = {'file_id': '1234'}

response = requests.put(url, files=files, data=payload, verify=False)

请注意，您不需要显式地指定任何内容类型。

注:想评论上述答案之一，但不能因为低声誉，所以起草了一个新的回应在这里。

通过在POST请求中指定files参数，请求的Content-Type被自动设置为multipart/form-data(后面跟着用于分隔multipart负载中的每个主体部分的边界字符串)，无论您只发送文件，还是同时发送form-data和文件(因此，在这种情况下，不应该尝试手动设置Content-Type)。然而，如果只发送表单数据，则Content-Type将自动设置为application/x-www-form-urlencoded。

You can print out the Content-Type header of the request to verify the above using the example given below, which shows how to upload multiple files (or a single file) with (optionally) the same key (i.e., 'files' in the case below), as well as with optional form-data (i.e., data=data in the example below). The documentation on how to POST single and multiple files can be found here and here, respectively. In case you need to upload large files without reading them into memory, have a look at Streaming Uploads. For the server side—in case this is needed—please have a look at this answer, from which the code snippet below has been taken, and which uses the FastAPI web framework.

import requests

url = 'http://127.0.0.1:8000/submit'
files = [('files', open('a.txt', 'rb')), ('files', open('b.txt', 'rb'))]
#file = {'file': open('a.txt','rb')} # to send a single file
data ={"name": "foo", "point": 0.13, "is_accepted": False}
r = requests.post(url=url, data=data, files=files) 
print(r.json())
print(r.request.headers['content-type'])