If upload_file is meant to be the file, use:

files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}

r = requests.post(url, files=files, data=values)

and requests will send a multi-part form POST body with the upload_file field set to the contents of the file.txt file.

The filename will be included in the mime header for the specific field:

>>> import requests
>>> open('file.txt', 'wb')  # create an empty demo file
<_io.BufferedWriter name='file.txt'>
>>> files = {'upload_file': open('file.txt', 'rb')}
>>> print(requests.Request('POST', 'http://example.com', files=files).prepare().body.decode('ascii'))
--c226ce13d09842658ffbd31e0563c6bd
Content-Disposition: form-data; name="upload_file"; filename="file.txt"


--c226ce13d09842658ffbd31e0563c6bd--

Note the filename="file.txt" parameter.

You can use a tuple for the files mapping value, with between 2 and 4 elements, if you need more control. The first element is the filename, followed by the contents, and an optional content-type header value and an optional mapping of additional headers:

files = {'upload_file': ('foobar.txt', open('file.txt','rb'), 'text/x-spam')}

This sets an alternative filename and content type, leaving out the optional headers.

If you are meaning the whole POST body to be taken from a file (with no other fields specified), then don't use the files parameter, just post the file directly as data. You then may want to set a Content-Type header too, as none will be set otherwise. See Python requests - POST data from a file.

A few points :

• make sure to submit your request to the correct url ( the form 'action' )
• use the data parameter to submit other form fields ( 'dir', 'submit' )
• include the name of the file in files ( this is optional )

code :

import requests

url = 'http://example.com' + '/upload.php'
data = {'dir':'/uploads/', 'submit':'Submit'}
files = {'file':('1.jpg', open('1.jpg', 'rb'))}
r = requests.post(url, data=data, files=files)

print(r.content)