This is my code for uploading to google drive with python requests using google-drive-api.
import sys
import json
import requests
from tqdm import tqdm
import requests_toolbelt
from requests.exceptions import JSONDecodeError
class ProgressBar(tqdm):
def update_to(self, n: int) -> None:
self.update(n - self.n)
def upload_file(access_token:str, filename:str, filedirectory:str):
metadata = {
"title": filename,
}
files = {}
session = requests.session()
with open(filedirectory, "rb") as fp:
files["file"] = fp
files["data"] = ('metadata', json.dumps(metadata), 'application/json')
encoder = requests_toolbelt.MultipartEncoder(files)
with ProgressBar(
total=encoder.len,
unit="B",
unit_scale=True,
unit_divisor=1024,
miniters=1,
file=sys.stdout,
) as bar:
monitor = requests_toolbelt.MultipartEncoderMonitor(
encoder, lambda monitor: bar.update_to(monitor.bytes_read)
)
r = session.post(
"https://www.googleapis.com/upload/drive/v3/files?uploadType=multipart",
data=monitor,
allow_redirects=False,
headers={"Authorization": "Bearer " + access_token},
)
try:
resp = r.json()
print(resp)
except JSONDecodeError:
sys.exit(r.text)
upload_file("access_token", "test.txt", "test.txt")
When i am trying send file with data attribute in post request then file name did not send and with files attribute in post request then requests-toolbelt not working. How to fix this error ?
When I saw your script, I thought that the content type is not included in the request header. In this case, I think that the request body is directly shown in the uploaded file. I thought that this might be the reason for your current issue. In order to remove this issue, how about the following modification?
From:
r = session.post(
url,
data=monitor,
allow_redirects=False,
headers={"Authorization": "Bearer " + access_token},
)
To:
r = session.post(
url,
data=monitor,
allow_redirects=False,
headers={
"Authorization": "Bearer " + access_token,
"Content-Type": monitor.content_type,
},
)
In this case, from metadata = { "title": filename }, it supposes that url is https://www.googleapis.com/upload/drive/v2/files?uploadType=multipart. Please be careful about this.
When you want to use Drive API v3, please modify metadata = { "title": filename } to metadata = { "name": filename }, and use the endpoint of https://www.googleapis.com/upload/drive/v3/files?uploadType=multipart.
When the file is uploaded with Drive API v3, the value of {'kind': 'drive#file', 'id': '###', 'name': 'test.txt', 'mimeType': 'text/plain'} is returned.
By the way, when an error like badContent occurs in your testing, please try to test the following modification. When in the request body of multipart/form-data the file content is put before the file metadata, it seems that an error occurs. I'm not sure whether this is the current specification. But, I didn't know the order of request body is required to be checked.
From
files = {}
files["file"] = fp
files["data"] = ('metadata', json.dumps(metadata), 'application/json')
To
files = collections.OrderedDict(data=("metadata", json.dumps(metadata), "application/json"), file=fp)
Note:
I thought that in your script, an error might occur at file_size = os.path.getsize(filename). Please confirm this again.
When I tested your script by modifying the above modifications, I could confirm that a test file could be uploaded to Google Drive with the expected filename. In this case, I also modified it as follows.
files = collections.OrderedDict(data=("metadata", json.dumps(metadata), "application/json"), file=fp)
References:
Files: insert of Drive API v2
Files: create of Drive API v3
Upload file data
Metadata needs to be sent in the post body as json.
Python Requests post() Method
data = Optional. A dictionary, list of tuples, bytes or a file object to send to the specified url
json = Optional. A JSON object to send to the specified url
metadata = {
"name": filename,
}
r = session.post(
url,
json=json.dumps(metadata),
allow_redirects=False,
headers={"Authorization": "Bearer " + access_token},
)
Future readers can find below a complete script that also contains details on how to get access to the bearer token for HTTP authentication.
Most of the credit goes to the OP and answers to the OPs question.
"""
Goal: For one time upload of a large file (as the GDrive UI hangs up)
Step 1 - Create OAuth 2.0 Client ID + Client Secret
- by following the "Authentication" part of https://pythonhosted.org/PyDrive/quickstart.html
Step 2 - Get Access Token
- from the OAuth playground -> https://developers.google.com/oauthplayground/
--> Select Drive API v3 -> www.googleapis.com/auth/drive --> Click on "Authorize APIs"
--> Click on "Exchange authorization code for tokens" --> "Copy paste the access token"
--> Use it in the script below
Step 3 - Run file as daemon process
- nohup python -u upload_gdrive.py > upload_gdrive.log 2>&1 &
- tail -f upload_gdrive.log
"""
import sys
import json
import requests
from tqdm import tqdm
import requests_toolbelt # pip install requests_toolbelt
from requests.exceptions import JSONDecodeError
import collections
class ProgressBar(tqdm):
def update_to(self, n: int) -> None:
self.update(n - self.n)
def upload_file(access_token:str, filename:str, filepath:str):
metadata = {
"name": filename,
}
files = {}
session = requests.session()
with open(filepath, "rb") as fp:
files = collections.OrderedDict(data=("metadata", json.dumps(metadata), "application/json"), file=fp)
encoder = requests_toolbelt.MultipartEncoder(files)
with ProgressBar(
total=encoder.len,
unit="B",
unit_scale=True,
unit_divisor=1024,
miniters=1,
file=sys.stdout,
) as bar:
monitor = requests_toolbelt.MultipartEncoderMonitor(
encoder, lambda monitor: bar.update_to(monitor.bytes_read)
)
r = session.post(
"https://www.googleapis.com/upload/drive/v3/files?uploadType=multipart",
data=monitor,
allow_redirects=False,
headers={
"Authorization": "Bearer " + access_token
, "Content-Type": monitor.content_type
},
)
try:
resp = r.json()
print(resp)
except JSONDecodeError:
sys.exit(r.text)
upload_file("<access_token>"
, "<upload_filename>", "<path_to_file>")
How to send a multipart/form-data with requests in python? How to send a file, I understand, but how to send the form data by this method can not understand.
Basically, if you specify a files parameter (a dictionary), then requests will send a multipart/form-data POST instead of a application/x-www-form-urlencoded POST. You are not limited to using actual files in that dictionary, however:
>>> import requests
>>> response = requests.post('http://httpbin.org/post', files=dict(foo='bar'))
>>> response.status_code
200
and httpbin.org lets you know what headers you posted with; in response.json() we have:
>>> from pprint import pprint
>>> pprint(response.json()['headers'])
{'Accept': '*/*',
'Accept-Encoding': 'gzip, deflate',
'Connection': 'close',
'Content-Length': '141',
'Content-Type': 'multipart/form-data; '
'boundary=c7cbfdd911b4e720f1dd8f479c50bc7f',
'Host': 'httpbin.org',
'User-Agent': 'python-requests/2.21.0'}
Better still, you can further control the filename, content type and additional headers for each part by using a tuple instead of a single string or bytes object. The tuple is expected to contain between 2 and 4 elements; the filename, the content, optionally a content type, and an optional dictionary of further headers.
I'd use the tuple form with None as the filename, so that the filename="..." parameter is dropped from the request for those parts:
>>> files = {'foo': 'bar'}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--bb3f05a247b43eede27a124ef8b968c5
Content-Disposition: form-data; name="foo"; filename="foo"
bar
--bb3f05a247b43eede27a124ef8b968c5--
>>> files = {'foo': (None, 'bar')}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--d5ca8c90a869c5ae31f70fa3ddb23c76
Content-Disposition: form-data; name="foo"
bar
--d5ca8c90a869c5ae31f70fa3ddb23c76--
files can also be a list of two-value tuples, if you need ordering and/or multiple fields with the same name:
requests.post(
'http://requestb.in/xucj9exu',
files=(
('foo', (None, 'bar')),
('foo', (None, 'baz')),
('spam', (None, 'eggs')),
)
)
If you specify both files and data, then it depends on the value of data what will be used to create the POST body. If data is a string, only it willl be used; otherwise both data and files are used, with the elements in data listed first.
There is also the excellent requests-toolbelt project, which includes advanced Multipart support. It takes field definitions in the same format as the files parameter, but unlike requests, it defaults to not setting a filename parameter. In addition, it can stream the request from open file objects, where requests will first construct the request body in memory:
from requests_toolbelt.multipart.encoder import MultipartEncoder
mp_encoder = MultipartEncoder(
fields={
'foo': 'bar',
# plain file object, no filename or mime type produces a
# Content-Disposition header with just the part name
'spam': ('spam.txt', open('spam.txt', 'rb'), 'text/plain'),
}
)
r = requests.post(
'http://httpbin.org/post',
data=mp_encoder, # The MultipartEncoder is posted as data, don't use files=...!
# The MultipartEncoder provides the content-type header with the boundary:
headers={'Content-Type': mp_encoder.content_type}
)
Fields follow the same conventions; use a tuple with between 2 and 4 elements to add a filename, part mime-type or extra headers. Unlike the files parameter, no attempt is made to find a default filename value if you don't use a tuple.
Requests has changed since some of the previous answers were written. Have a look at this Issue on Github for more details and this comment for an example.
In short, the files parameter takes a dictionary with the key being the name of the form field and the value being either a string or a 2, 3 or 4-length tuple, as described in the section POST a Multipart-Encoded File in the Requests quickstart:
>>> url = 'http://httpbin.org/post'
>>> files = {'file': ('report.xls', open('report.xls', 'rb'), 'application/vnd.ms-excel', {'Expires': '0'})}
In the above, the tuple is composed as follows:
(filename, data, content_type, headers)
If the value is just a string, the filename will be the same as the key, as in the following:
>>> files = {'obvius_session_id': '72c2b6f406cdabd578c5fd7598557c52'}
Content-Disposition: form-data; name="obvius_session_id"; filename="obvius_session_id"
Content-Type: application/octet-stream
72c2b6f406cdabd578c5fd7598557c52
If the value is a tuple and the first entry is None the filename property will not be included:
>>> files = {'obvius_session_id': (None, '72c2b6f406cdabd578c5fd7598557c52')}
Content-Disposition: form-data; name="obvius_session_id"
Content-Type: application/octet-stream
72c2b6f406cdabd578c5fd7598557c52
You need to use the files parameter to send a multipart form POST request even when you do not need to upload any files.
From the original requests source:
def request(method, url, **kwargs):
"""Constructs and sends a :class:`Request <Request>`.
...
:param files: (optional) Dictionary of ``'name': file-like-objects``
(or ``{'name': file-tuple}``) for multipart encoding upload.
``file-tuple`` can be a 2-tuple ``('filename', fileobj)``,
3-tuple ``('filename', fileobj, 'content_type')``
or a 4-tuple ``('filename', fileobj, 'content_type', custom_headers)``,
where ``'content-type'`` is a string
defining the content type of the given file
and ``custom_headers`` a dict-like object
containing additional headers to add for the file.
The relevant part is: file-tuple can be a:
2-tuple (filename, fileobj)
3-tuple (filename, fileobj, content_type)
4-tuple (filename, fileobj, content_type, custom_headers).
☝ What might not be obvious is that fileobj can be either an actual file object when dealing with files, OR a string when dealing with plain text fields.
Based on the above, the simplest multipart form request that includes both files to upload and form fields will look like this:
import requests
multipart_form_data = {
'upload': ('custom_file_name.zip', open('myfile.zip', 'rb')),
'action': (None, 'store'),
'path': (None, '/path1')
}
response = requests.post('https://httpbin.org/post', files=multipart_form_data)
print(response.content)
☝ Note the None as the first argument in the tuple for plain text fields — this is a placeholder for the filename field which is only used for file uploads, but for text fields passing None as the first parameter is required in order for the data to be submitted.
Multiple fields with the same name
If you need to post multiple fields with the same name then instead of a dictionary you can define your payload as a list (or a tuple) of tuples:
multipart_form_data = (
('file2', ('custom_file_name.zip', open('myfile.zip', 'rb'))),
('action', (None, 'store')),
('path', (None, '/path1')),
('path', (None, '/path2')),
('path', (None, '/path3')),
)
Streaming requests API
If the above API is not pythonic enough for you, then consider using requests toolbelt (pip install requests_toolbelt) which is an extension of the core requests module that provides support for file upload streaming as well as the MultipartEncoder which can be used instead of files, and which also lets you define the payload as a dictionary, tuple or list.
MultipartEncoder can be used both for multipart requests with or without actual upload fields. It must be assigned to the data parameter.
import requests
from requests_toolbelt.multipart.encoder import MultipartEncoder
multipart_data = MultipartEncoder(
fields={
# a file upload field
'file': ('file.zip', open('file.zip', 'rb'), 'text/plain')
# plain text fields
'field0': 'value0',
'field1': 'value1',
}
)
response = requests.post('http://httpbin.org/post', data=multipart_data,
headers={'Content-Type': multipart_data.content_type})
If you need to send multiple fields with the same name, or if the order of form fields is important, then a tuple or a list can be used instead of a dictionary:
multipart_data = MultipartEncoder(
fields=(
('action', 'ingest'),
('item', 'spam'),
('item', 'sausage'),
('item', 'eggs'),
)
)
Here is the simple code snippet to upload a single file with additional parameters using requests:
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)
Please note that you don't need to explicitly specify any content type.
NOTE: Wanted to comment on one of the above answers but could not because of low reputation so drafted a new response here.
By specifying a files parameter in the POST request, the Content-Type of the request is automatically set to multipart/form-data (followed by the boundary string used to separate each body part in the multipart payload), whether you send only files, or form-data and files at the same time (thus, one shouldn't attempt setting the Content-Type manually in this case). Whereas, if only form-data were sent, the Content-Type would automatically be set to 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'])
You need to use the name attribute of the upload file that is in the HTML of the site. Example:
autocomplete="off" name="image">
You see name="image">? You can find it in the HTML of a site for uploading the file. You need to use it to upload the file with Multipart/form-data
script:
import requests
site = 'https://prnt.sc/upload.php' # the site where you upload the file
filename = 'image.jpg' # name example
Here, in the place of image, add the name of the upload file in HTML
up = {'image':(filename, open(filename, 'rb'), "multipart/form-data")}
If the upload requires to click the button for upload, you can use like that:
data = {
"Button" : "Submit",
}
Then start the request
request = requests.post(site, files=up, data=data)
And done, file uploaded succesfully
import requests
# assume sending two files
url = "put ur url here"
f1 = open("file 1 path", 'rb')
f2 = open("file 2 path", 'rb')
response = requests.post(url,files={"file1 name": f1, "file2 name":f2})
print(response)
To clarify examples given above,
"You need to use the files parameter to send a multipart form POST request even when you do not need to upload any files."
files={}
won't work, unfortunately.
You will need to put some dummy values in, e.g.
files={"foo": "bar"}
I came up against this when trying to upload files to Bitbucket's REST API and had to write this abomination to avoid the dreaded "Unsupported Media Type" error:
url = "https://my-bitbucket.com/rest/api/latest/projects/FOO/repos/bar/browse/foobar.txt"
payload = {'branch': 'master',
'content': 'text that will appear in my file',
'message': 'uploading directly from python'}
files = {"foo": "bar"}
response = requests.put(url, data=payload, files=files)
:O=
Send multipart/form-data key and value
curl command:
curl -X PUT http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F taskStatus=1
python requests - More complicated POST requests:
updateTaskUrl = "http://127.0.0.1:8080/api/xxx"
updateInfoDict = {
"taskStatus": 1,
}
resp = requests.put(updateTaskUrl, data=updateInfoDict)
Send multipart/form-data file
curl command:
curl -X POST http://127.0.0.1:8080/api/xxx ...
-H 'content-type: multipart/form-data; boundary=----xxx' \
-F file=#/Users/xxx.txt
python requests - POST a Multipart-Encoded File:
filePath = "/Users/xxx.txt"
fileFp = open(filePath, 'rb')
fileInfoDict = {
"file": fileFp,
}
resp = requests.post(uploadResultUrl, files=fileInfoDict)
that's all.
import json
import os
import requests
from requests_toolbelt import MultipartEncoder
AUTH_API_ENDPOINT = "http://localhost:3095/api/auth/login"
def file_upload(path_img, token ):
url = 'http://localhost:3095/api/shopping/product/image'
name_img = os.path.basename(path_img)
mp_encoder = MultipartEncoder(
fields={
'email': 'mcm9#gmail.com',
'source': 'tmall',
'productId': 'product_0001',
'image': (name_img, open(path_img, 'rb'), 'multipart/form-data')
#'spam': ('spam.txt', open('spam.txt', 'rb'), 'text/plain'),
}
)
head = {'Authorization': 'Bearer {}'.format(token),
'Content-Type': mp_encoder.content_type}
with requests.Session() as s:
result = s.post(url, data=mp_encoder, headers=head)
return result
def do_auth(username, password, url=AUTH_API_ENDPOINT):
data = {
"email": username,
"password": password
}
# sending post request and saving response as response object
r = requests.post(url=url, data=data)
# extracting response text
response_text = r.text
d = json.loads(response_text)
# print(d)
return d
if __name__ == '__main__':
result = do_auth('mcm4#gmail.com','123456')
token = result.get('data').get('payload').get('token')
print(token)
result = file_upload('/home/mcm/Pictures/1234.png',token)
print(result.json())
Here is the python snippet you need to upload one large single file as multipart formdata. With NodeJs Multer middleware running on the server side.
import requests
latest_file = 'path/to/file'
url = "http://httpbin.org/apiToUpload"
files = {'fieldName': open(latest_file, 'rb')}
r = requests.put(url, files=files)
For the server side please check the multer documentation at: https://github.com/expressjs/multer
here the field single('fieldName') is used to accept one single file, as in:
var upload = multer().single('fieldName');
This is one way to send file in multipart request
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)
Other approach
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)
I have tested both , both works fine.
I'm trying to send a request to URL_server with request module in python 3.
This works for me:
# -*- 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()
Where:
URL_SERVER_TO_POST_DATA = Server where we going to send data
HEADERS = Headers sended
file_data = Params sended
Postman generated code for file upload with additional form fields:
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"))
I need to do a API call to upload a file along with a JSON string with details about the file.
I am trying to use the python requests lib to do this:
import requests
info = {
'var1' : 'this',
'var2' : 'that',
}
data = json.dumps({
'token' : auth_token,
'info' : info,
})
headers = {'Content-type': 'multipart/form-data'}
files = {'document': open('file_name.pdf', 'rb')}
r = requests.post(url, files=files, data=data, headers=headers)
This throws the following error:
raise ValueError("Data must not be a string.")
ValueError: Data must not be a string
If I remove the 'files' from the request, it works.
If I remove the 'data' from the request, it works.
If I do not encode data as JSON it works.
For this reason I think the error is to do with sending JSON data and files in the same request.
Any ideas on how to get this working?
See this thread How to send JSON as part of multipart POST-request
Do not set the Content-type header yourself, leave that to pyrequests to generate
def send_request():
payload = {"param_1": "value_1", "param_2": "value_2"}
files = {
'json': (None, json.dumps(payload), 'application/json'),
'file': (os.path.basename(file), open(file, 'rb'), 'application/octet-stream')
}
r = requests.post(url, files=files)
print(r.content)
Don't encode using json.
import requests
info = {
'var1' : 'this',
'var2' : 'that',
}
data = {
'token' : auth_token,
'info' : info,
}
headers = {'Content-type': 'multipart/form-data'}
files = {'document': open('file_name.pdf', 'rb')}
r = requests.post(url, files=files, data=data, headers=headers)
Note that this may not necessarily be what you want, as it will become another form-data section.
I'm don't think you can send both data and files in a multipart encoded file, so you need to make your data a "file" too:
files = {
'data' : data,
'document': open('file_name.pdf', 'rb')
}
r = requests.post(url, files=files, headers=headers)
I have been using requests==2.22.0
For me , the below code worked.
import requests
data = {
'var1': 'this',
'var2': 'that'
}
r = requests.post("http://api.example.com/v1/api/some/",
files={'document': open('doocument.pdf', 'rb')},
data=data,
headers={"Authorization": "Token jfhgfgsdadhfghfgvgjhN"}. #since I had to authenticate for the same
)
print (r.json())
For sending Facebook Messenger API, I changed all the payload dictionary values to be strings. Then, I can pass the payload as data parameter.
import requests
ACCESS_TOKEN = ''
url = 'https://graph.facebook.com/v2.6/me/messages'
payload = {
'access_token' : ACCESS_TOKEN,
'messaging_type' : "UPDATE",
'recipient' : '{"id":"1111111111111"}',
'message' : '{"attachment":{"type":"image", "payload":{"is_reusable":true}}}',
}
files = {'filedata': (file, open(file, 'rb'), 'image/png')}
r = requests.post(url, files=files, data=payload)
1. Sending request
import json
import requests
cover = 'superneat.jpg'
payload = {'title': 'The 100 (2014)', 'episodes': json.dumps(_episodes)}
files = [
('json', ('payload.json', json.dumps(payload), 'application/json')),
('cover', (cover, open(cover, 'rb')))
]
r = requests.post("https://superneatech.com/store/series", files=files)
print(r.text)
2. Receiving request
You will receive the JSON data as a file, get the content and continue...
Reference: View Here
What is more:
files = {
'document': open('file_name.pdf', 'rb')
}
That will only work if your file is at the same directory where your script is.
If you want to append file from different directory you should do:
files = {
'document': open(os.path.join(dir_path, 'file_name.pdf'), 'rb')
}
Where dir_path is a directory with your 'file_name.pdf' file.
But what if you'd like to send multiple PDFs ?
You can simply make a custom function to return a list of files you need (in your case that can be only those with .pdf extension). That also includes files in subdirectories (search for files recursively):
def prepare_pdfs():
return sorted([os.path.join(root, filename) for root, dirnames, filenames in os.walk(dir_path) for filename in filenames if filename.endswith('.pdf')])
Then you can call it:
my_data = prepare_pdfs()
And with simple loop:
for file in my_data:
pdf = open(file, 'rb')
files = {
'document': pdf
}
r = requests.post(url, files=files, ...)