How to Send a "Multipart/Form-Data" With Requests in Python

How to POST multipart/form-data in Python with requests

try with this

import requests

headers = {
    'Content-Type': 'multipart/form-data',
    'x-api-key': 'xxxxxx-xxxxx-xxxx-xxxx-xxxxxxxx',
}

files = {
    'file': ('<image>.jpg', open('<image>.jpg', 'rb')),
}

response = requests.post('http://localhost:8000/api/v1/recognition/recognize', headers=headers, files=files)

Upload file via POST multipart form-data using Python

Just post it without stream bytes and header magic:

file = {'BackupFile': open(file_name, 'rb')}
response = requests.post(url, files=file)

all the rest will make requests: will send a multi-part form POST body with the BackupFile field set to the contents of the file.txt file.
Also the filename will be included in the mime header for the specific field.

example:

>>> files = {'BackupFile': open('file.txt', 'rb')}
>>> print(requests.Request('POST', 'http://example.com', files=files).prepare().body.decode('ascii'))
--0c249de0e6243e92307003732e49ffe9
Content-Disposition: form-data; name="BackupFile"; filename="file.txt"
--0c249de0e6243e92307003732e49ffe9--

How to send a 'multipart/form-data' with requests, without files in Python

I'm not entirely sure how python works with requests. You can try this accepted answers or any which is suitable

python requests - POST Multipart/form-data without filename in HTTP request

requests: post multipart/form-data

Removed 'content-type' from the headers, now it works

try:
    headers = {
        'authorization': "Basic ZXNlbnRpcmVcY       


       
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dddddd",
        'cache-control': "no-cache",
    }
    myfile = {"file": ("filexxx", open(filepath, "rb"))}

    response = requests.request("POST", verify=False, url=url, data=myfile, headers=headers)

    print(response.text)
except Exception as e:
    print "Exception:", e

How to send a “multipart/form-data” with requests in python?

payload = { 'username':'xxxxx',
            'passwd':'xxxx'}
session = requests.Session()
req = session.post('https://exemple.com/0',data=payload)

payload ={'check_type_diff':'0',
          'category':'19',
          'company_ad':'0'}


req = session.post('https://exemple.com/0', data=payload )
print req.content

Note: You should use post('URL',files=files) if you have file content. The multipart data just works as a normal data, just the formatting and the method is not the same.

Example:
If you have a file and some multipart data, your code will be like this:

files = {"file":(filename1,open(location+'/'+filename,"rb"),'application-type')}
payload ={'file-name':'Filename',
          'category':'19'}


req = session.post('https://exemple.com/0', data=payload, file=files)
print req.content

You don't even need to add the line "file" into the payload, the requests will put the request together.

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