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I want to send image from my computer to server [duplicate]

I'm performing a simple task of uploading a file using Python requests library. I searched Stack Overflow and no one seemed to have the same problem, namely, that the file is not received by the server:
import requests
url='http://nesssi.cacr.caltech.edu/cgi-bin/getmulticonedb_release2.cgi/post'
files={'files': open('file.txt','rb')}
values={'upload_file' : 'file.txt' , 'DB':'photcat' , 'OUT':'csv' , 'SHORT':'short'}
r=requests.post(url,files=files,data=values)
I'm filling the value of 'upload_file' keyword with my filename, because if I leave it blank, it says
Error - You must select a file to upload!
And now I get
File file.txt of size bytes is uploaded successfully!
Query service results: There were 0 lines.
Which comes up only if the file is empty. So I'm stuck as to how to send my file successfully. I know that the file works because if I go to this website and manually fill in the form it returns a nice list of matched objects, which is what I'm after. I'd really appreciate all hints.
Some other threads related (but not answering my problem):
Send file using POST from a Python script
http://docs.python-requests.org/en/latest/user/quickstart/#response-content
Uploading files using requests and send extra data
http://docs.python-requests.org/en/latest/user/advanced/#body-content-workflow

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.

(2018) the new python requests library has simplified this process, we can use the 'files' variable to signal that we want to upload a multipart-encoded file
url = 'http://httpbin.org/post'
files = {'file': open('report.xls', 'rb')}
r = requests.post(url, files=files)
r.text

Client Upload
If you want to upload a single file with Python requests library, then requests lib supports streaming uploads, which allow you to send large files or streams without reading into memory.
with open('massive-body', 'rb') as f:
requests.post('http://some.url/streamed', data=f)
Server Side
Then store the file on the server.py side such that save the stream into file without loading into the memory. Following is an example with using Flask file uploads.
#app.route("/upload", methods=['POST'])
def upload_file():
from werkzeug.datastructures import FileStorage
FileStorage(request.stream).save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return 'OK', 200
Or use werkzeug Form Data Parsing as mentioned in a fix for the issue of "large file uploads eating up memory" in order to avoid using memory inefficiently on large files upload (s.t. 22 GiB file in ~60 seconds. Memory usage is constant at about 13 MiB.).
#app.route("/upload", methods=['POST'])
def upload_file():
def custom_stream_factory(total_content_length, filename, content_type, content_length=None):
import tempfile
tmpfile = tempfile.NamedTemporaryFile('wb+', prefix='flaskapp', suffix='.nc')
app.logger.info("start receiving file ... filename => " + str(tmpfile.name))
return tmpfile
import werkzeug, flask
stream, form, files = werkzeug.formparser.parse_form_data(flask.request.environ, stream_factory=custom_stream_factory)
for fil in files.values():
app.logger.info(" ".join(["saved form name", fil.name, "submitted as", fil.filename, "to temporary file", fil.stream.name]))
# Do whatever with stored file at `fil.stream.name`
return 'OK', 200

You can send any file via post api while calling the API just need to mention files={'any_key': fobj}
import requests
import json
url = "https://request-url.com"
headers = {"Content-Type": "application/json; charset=utf-8"}
with open(filepath, 'rb') as fobj:
response = requests.post(url, headers=headers, files={'file': fobj})
print("Status Code", response.status_code)
print("JSON Response ", response.json())

#martijn-pieters answer is correct, however I wanted to add a bit of context to data= and also to the other side, in the Flask server, in the case where you are trying to upload files and a JSON.
From the request side, this works as Martijn describes:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
However, on the Flask side (the receiving webserver on the other side of this POST), I had to use form
#app.route("/sftp-upload", methods=["POST"])
def upload_file():
if request.method == "POST":
# the mimetype here isnt application/json
# see here: https://stackoverflow.com/questions/20001229/how-to-get-posted-json-in-flask
body = request.form
print(body) # <- immutable dict
body = request.get_json() will return nothing. body = request.get_data() will return a blob containing lots of things like the filename etc.
Here's the bad part: on the client side, changing data={} to json={} results in this server not being able to read the KV pairs! As in, this will result in a {} body above:
r = requests.post(url, files=files, json=values). # No!
This is bad because the server does not have control over how the user formats the request; and json= is going to be the habbit of requests users.

Upload:
with open('file.txt', 'rb') as f:
files = {'upload_file': f.read()}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
Download (Django):
with open('file.txt', 'wb') as f:
f.write(request.FILES['upload_file'].file.read())

Regarding the answers given so far, there was always something missing that prevented it to work on my side. So let me show you what worked for me:
import json
import os
import requests
API_ENDPOINT = "http://localhost:80"
access_token = "sdfJHKsdfjJKHKJsdfJKHJKysdfJKHsdfJKHs" # TODO: get fresh Token here
def upload_engagement_file(filepath):
url = API_ENDPOINT + "/api/files" # add any URL parameters if needed
hdr = {"Authorization": "Bearer %s" % access_token}
with open(filepath, "rb") as fobj:
file_obj = fobj.read()
file_basename = os.path.basename(filepath)
file_to_upload = {"file": (str(file_basename), file_obj)}
finfo = {"fullPath": filepath}
upload_response = requests.post(url, headers=hdr, files=file_to_upload, data=finfo)
fobj.close()
# print("Status Code ", upload_response.status_code)
# print("JSON Response ", upload_response.json())
return upload_response
Note that requests.post(...) needs
a url parameter, containing the full URL of the API endpoint you're calling, using the API_ENDPOINT, assuming we have an http://localhost:8000/api/files endpoint to POST a file
a headers parameter, containing at least the authorization (bearer token)
a files parameter taking the name of the file plus the entire file content
a data parameter taking just the path and file name
Installation required (console):
pip install requests
What you get back from the function call is a response object containing a status code and also the full error message in JSON format. The commented print statements at the end of upload_engagement_file are showing you how you can access them.
Note: Some useful additional information about the requests library can be found here

Some may need to upload via a put request and this is slightly different that posting data. It is important to understand how the server expects the data in order to form a valid request. A frequent source of confusion is sending multipart-form data when it isn't accepted. This example uses basic auth and updates an image via a put request.
url = 'foobar.com/api/image-1'
basic = requests.auth.HTTPBasicAuth('someuser', 'password123')
# Setting the appropriate header is important and will vary based
# on what you upload
headers = {'Content-Type': 'image/png'}
with open('image-1.png', 'rb') as img_1:
r = requests.put(url, auth=basic, data=img_1, headers=headers)
While the requests library makes working with http requests a lot easier, some of its magic and convenience obscures just how to craft more nuanced requests.

In Ubuntu you can apply this way,
to save file at some location (temporary) and then open and send it to API
path = default_storage.save('static/tmp/' + f1.name, ContentFile(f1.read()))
path12 = os.path.join(os.getcwd(), "static/tmp/" + f1.name)
data={} #can be anything u want to pass along with File
file1 = open(path12, 'rb')
header = {"Content-Disposition": "attachment; filename=" + f1.name, "Authorization": "JWT " + token}
res= requests.post(url,data,header)

how to upload a file using request.Post in python [duplicate]

I'm performing a simple task of uploading a file using Python requests library. I searched Stack Overflow and no one seemed to have the same problem, namely, that the file is not received by the server:
import requests
url='http://nesssi.cacr.caltech.edu/cgi-bin/getmulticonedb_release2.cgi/post'
files={'files': open('file.txt','rb')}
values={'upload_file' : 'file.txt' , 'DB':'photcat' , 'OUT':'csv' , 'SHORT':'short'}
r=requests.post(url,files=files,data=values)
I'm filling the value of 'upload_file' keyword with my filename, because if I leave it blank, it says
Error - You must select a file to upload!
And now I get
File file.txt of size bytes is uploaded successfully!
Query service results: There were 0 lines.
Which comes up only if the file is empty. So I'm stuck as to how to send my file successfully. I know that the file works because if I go to this website and manually fill in the form it returns a nice list of matched objects, which is what I'm after. I'd really appreciate all hints.
Some other threads related (but not answering my problem):
Send file using POST from a Python script
http://docs.python-requests.org/en/latest/user/quickstart/#response-content
Uploading files using requests and send extra data
http://docs.python-requests.org/en/latest/user/advanced/#body-content-workflow

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.

(2018) the new python requests library has simplified this process, we can use the 'files' variable to signal that we want to upload a multipart-encoded file
url = 'http://httpbin.org/post'
files = {'file': open('report.xls', 'rb')}
r = requests.post(url, files=files)
r.text

Client Upload
If you want to upload a single file with Python requests library, then requests lib supports streaming uploads, which allow you to send large files or streams without reading into memory.
with open('massive-body', 'rb') as f:
requests.post('http://some.url/streamed', data=f)
Server Side
Then store the file on the server.py side such that save the stream into file without loading into the memory. Following is an example with using Flask file uploads.
#app.route("/upload", methods=['POST'])
def upload_file():
from werkzeug.datastructures import FileStorage
FileStorage(request.stream).save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return 'OK', 200
Or use werkzeug Form Data Parsing as mentioned in a fix for the issue of "large file uploads eating up memory" in order to avoid using memory inefficiently on large files upload (s.t. 22 GiB file in ~60 seconds. Memory usage is constant at about 13 MiB.).
#app.route("/upload", methods=['POST'])
def upload_file():
def custom_stream_factory(total_content_length, filename, content_type, content_length=None):
import tempfile
tmpfile = tempfile.NamedTemporaryFile('wb+', prefix='flaskapp', suffix='.nc')
app.logger.info("start receiving file ... filename => " + str(tmpfile.name))
return tmpfile
import werkzeug, flask
stream, form, files = werkzeug.formparser.parse_form_data(flask.request.environ, stream_factory=custom_stream_factory)
for fil in files.values():
app.logger.info(" ".join(["saved form name", fil.name, "submitted as", fil.filename, "to temporary file", fil.stream.name]))
# Do whatever with stored file at `fil.stream.name`
return 'OK', 200

You can send any file via post api while calling the API just need to mention files={'any_key': fobj}
import requests
import json
url = "https://request-url.com"
headers = {"Content-Type": "application/json; charset=utf-8"}
with open(filepath, 'rb') as fobj:
response = requests.post(url, headers=headers, files={'file': fobj})
print("Status Code", response.status_code)
print("JSON Response ", response.json())

#martijn-pieters answer is correct, however I wanted to add a bit of context to data= and also to the other side, in the Flask server, in the case where you are trying to upload files and a JSON.
From the request side, this works as Martijn describes:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
However, on the Flask side (the receiving webserver on the other side of this POST), I had to use form
#app.route("/sftp-upload", methods=["POST"])
def upload_file():
if request.method == "POST":
# the mimetype here isnt application/json
# see here: https://stackoverflow.com/questions/20001229/how-to-get-posted-json-in-flask
body = request.form
print(body) # <- immutable dict
body = request.get_json() will return nothing. body = request.get_data() will return a blob containing lots of things like the filename etc.
Here's the bad part: on the client side, changing data={} to json={} results in this server not being able to read the KV pairs! As in, this will result in a {} body above:
r = requests.post(url, files=files, json=values). # No!
This is bad because the server does not have control over how the user formats the request; and json= is going to be the habbit of requests users.

Upload:
with open('file.txt', 'rb') as f:
files = {'upload_file': f.read()}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
Download (Django):
with open('file.txt', 'wb') as f:
f.write(request.FILES['upload_file'].file.read())

Regarding the answers given so far, there was always something missing that prevented it to work on my side. So let me show you what worked for me:
import json
import os
import requests
API_ENDPOINT = "http://localhost:80"
access_token = "sdfJHKsdfjJKHKJsdfJKHJKysdfJKHsdfJKHs" # TODO: get fresh Token here
def upload_engagement_file(filepath):
url = API_ENDPOINT + "/api/files" # add any URL parameters if needed
hdr = {"Authorization": "Bearer %s" % access_token}
with open(filepath, "rb") as fobj:
file_obj = fobj.read()
file_basename = os.path.basename(filepath)
file_to_upload = {"file": (str(file_basename), file_obj)}
finfo = {"fullPath": filepath}
upload_response = requests.post(url, headers=hdr, files=file_to_upload, data=finfo)
fobj.close()
# print("Status Code ", upload_response.status_code)
# print("JSON Response ", upload_response.json())
return upload_response
Note that requests.post(...) needs
a url parameter, containing the full URL of the API endpoint you're calling, using the API_ENDPOINT, assuming we have an http://localhost:8000/api/files endpoint to POST a file
a headers parameter, containing at least the authorization (bearer token)
a files parameter taking the name of the file plus the entire file content
a data parameter taking just the path and file name
Installation required (console):
pip install requests
What you get back from the function call is a response object containing a status code and also the full error message in JSON format. The commented print statements at the end of upload_engagement_file are showing you how you can access them.
Note: Some useful additional information about the requests library can be found here

Some may need to upload via a put request and this is slightly different that posting data. It is important to understand how the server expects the data in order to form a valid request. A frequent source of confusion is sending multipart-form data when it isn't accepted. This example uses basic auth and updates an image via a put request.
url = 'foobar.com/api/image-1'
basic = requests.auth.HTTPBasicAuth('someuser', 'password123')
# Setting the appropriate header is important and will vary based
# on what you upload
headers = {'Content-Type': 'image/png'}
with open('image-1.png', 'rb') as img_1:
r = requests.put(url, auth=basic, data=img_1, headers=headers)
While the requests library makes working with http requests a lot easier, some of its magic and convenience obscures just how to craft more nuanced requests.

In Ubuntu you can apply this way,
to save file at some location (temporary) and then open and send it to API
path = default_storage.save('static/tmp/' + f1.name, ContentFile(f1.read()))
path12 = os.path.join(os.getcwd(), "static/tmp/" + f1.name)
data={} #can be anything u want to pass along with File
file1 = open(path12, 'rb')
header = {"Content-Disposition": "attachment; filename=" + f1.name, "Authorization": "JWT " + token}
res= requests.post(url,data,header)

Python Falcon - get POST data

I try to use falcon package in my project. Problem is I didn't find a way to get body data from the HTTP post request.
I used code from example, but req.stream.read() doesn't return JSON as expected.
The code is:
raw_json = req.stream.read()
result.json(raw_json, encoding='utf-8')
resp.body = json.dumps(result_json, encoding='utf-8')
How to get the POST data?
Thanks for any help

in falcon 2, if you work with json type, use req.media
for example:
import falcon
from json import dumps
class Resource(object):
def on_post(self, req, resp, **kwargs):
result = req.media
# do your job
resp.body = dumps(result)
api = falcon.API()
api.add_route('/test', Resource())

Little digging into the problem led to the following linked issue on github. It states that falcon framework at least in its version 0.3 and working with Python 2 didn't parse data 'POSTed' as string if they are aptly escaped. We could use more information on what data you are trying to send over POST request and in what format is that being sent, as in if its being send as simple text, or with Header Information Content-Type:application/json, or if its coming through an HTML form.
While the exact issue is not clear from the question I could still suggest trying to use bounded_stream instead of stream as in:
raw_json = req.bounded_stream.read()
result.json(raw_json, encoding='utf-8')
resp.body = json.dumps(result_json, encoding='utf-8')
for the official documentation suggests use of bounded_stream where uncertain conditions such as Content-Length undefined or 0, or if header information is missing altogether.
bounded_stream is described as the following in the official falcon documentation.
File-like wrapper around stream to normalize certain differences between the native input objects employed by different WSGI servers. In particular, bounded_stream is aware of the expected Content-Length of the body, and will never block on out-of-bounds reads, assuming the client does not stall while transmitting the data to the server.
Falcon receives the HTTP requests data as buffer object as passed by WSGI wrapper which receives the data from client, and its possible it doesn't run proper parsing on top of the data to convert to a more usable data structure for performance reasons.

Big thanks to Ryan (and Prateek Jain) for the answer.
The solution is simply to put app.req_options.auto_parse_form_urlencoded=True. For example:
import falcon
class ThingsResource(object):
def on_post(self, req, resp):
value = req.get_param("value", required=True)
#do something with value
app = falcon.API()
app.req_options.auto_parse_form_urlencoded=True
things = ThingsResource()
app.add_route('/things', things)

The field you're looking for is somewhat confusingly named, but it's req.media:
Returns a deserialized form of the request stream. When called, it will attempt to deserialize the request stream using the Content-Type header as well as the media-type handlers configured via falcon.RequestOptions.
If the request is JSON, req.media already contains a python dict.

I have added changes in request.py in falcon framework to parse application/x-www-form-urlencoded and multipart/from-data.
I have raised pull request - https://github.com/falconry/falcon/pull/1236 but it is not yet merged in master.
Check this - https://github.com/branelmoro/falcon
I have added new code to parse POST, PUT and DELETE application/x-www-form-urlencoded and multipart/form-data.
Text fields will be available in req.form_data dictionary and upload file buffer stream will be available in req.files dictionary.
I hope this will help to access POST and GET parameters separately and we will be able to upload files as well.
Good thing about the change is that it will not load entire uploaded file in memory.
Below is sample code to show how to use POST, PUT and DELETE application/x-www-form-urlencoded and multipart/form-data:
import falcon
class Resource(object):
def on_post(self, req, resp):
# req.form_data will return dictionary of text field names and their values
print(req.form_data)
# req.form_data will return dictionary of file field names and
# their buffer class FileStream objects as values
print(req.files)
# support we are uploading a image.jpg in `pancard` file field then
# req.files["pancard"] will be FileStream buffer object
# We can use set_max_upload_size method to set maximum allowed
# file size let say 1Mb = 1*1024*1024 bytes for this file
req.files["pancard"].set_max_upload_size(1*1024*1024)
# We can use uploadto method to upload file on required path (Note: absolute filepath is required)
# This method returns boolean - `True` on successful upload
# and if upload is unsuccessful then it returns `False` and sets error on failure.
path = "/tmp/" + req.files["pancard"].name
response = req.files["pancard"].uploadto("/tmp/" + path)
print(response)
# Once file is uploaded sucessfully, we can check it's size
print(req.files["pancard"].size)
# If file is not uploaded sucessfully, we can check it's error
print(req.files["pancard"].error)
resp.body = "Done file upload"
resp.status = falcon.HTTP_200
# falcon.API instances are callable WSGI apps
app = falcon.API()
things = Resource()
# things will handle post requests to the '/post_path' URL path
app.add_route('/post_path', things)
Do let me know if you have any doubts.

So far... for me bounded_stream.read() and stream.read() both get the posted data as type str. I have only found one way around the issue so far:
def on_post(self, req, resp):
posted_data = json.loads(req.stream.read())
print(str(type(posted_data)))
print(posted_data)
Loading the string into a json dict once the posted data is received is my only solution that I can come up with

Here's something I used while designing an API.
import falcon
import json
class VerifierResource():
def on_post(self, req, resp):
credentials = json.loads(req.stream.read())
if credentials['username'] == USER \
and credentials['passwd'] == PASSWORD:
resp.body = json.dumps({"status": "verified"})
else:
resp.body = json.dumps({"status": "invalid"})
api = falcon.API()
api.add_route('/verify', VerifierResource())
This returns a serialized JSON with corresponding response body.

there is a sample way to get media from body. I use to get the body in the post method:
def on_post(req,resp)
arguments = {}
# get body media on post method
body = req.get_media()
if 'something' in body:
arguments['something'] = body['something']
send body content type Media-Type and print resp or use in code, but if want to send JSON body your code should cover give JSON parameters.
Do let me know if you have any doubts.

Flask Restful: change representation based on URL parameter

I am building an API using Flask and Flask-Restful. The API might be accessed by different sort of tools (web apps, automated tools, etc.) and one of the requirement is to provide different representations (let's say json and csv for the sake of the example)
As explained in the restful doc, it's easy to change the serialization based on the content type, so for my CSV serialization I've added this:
#api.representation('text/csv')
def output_csv(data, code, headers=None):
#some CSV serialized data
data = 'some,csv,fields'
resp = app.make_response(data)
return resp
And it's working when using curl and passing the correct -H "Accept: text/csv" parameter.
The issue is that since some browsers might be routed to a url directly to download a csv file, I would like to be able to force my serialization via a url parameter for example http://my.domain.net/api/resource?format=csv where the format=csvwould have the same effect as -H "Accept: text/csv".
I've gone through both Flask and Flask-Restful documentation and I don't see how to correctly handle this.

Simply create a sub-class of Api and override the mediatypes method:
from werkzeug.exceptions import NotAcceptable
class CustomApi(Api):
FORMAT_MIMETYPE_MAP = {
"csv": "text/csv",
"json": "application/json"
# Add other mimetypes as desired here
}
def mediatypes(self):
"""Allow all resources to have their representation
overriden by the `format` URL argument"""
preferred_response_type = []
format = request.args.get("format")
if format:
mimetype = FORMAT_MIMETYPE_MAP.get(format)
preferred_response_type.append(mimetype)
if not mimetype:
raise NotAcceptable()
return preferred_response_type + super(CustomApi, self).mediatypes()

Basically you want to retrieve parameters from GET method. Please refer to:
How do I get the url parameter in a Flask view

How can I post an attachment to jira via python

I want to use python to post an attachment to jira via jira rest api and without any other packages which are needed to install.
I noticed that "This resource expects a multipart post.", and I tried it,but maybe my method was wrong,I failed
I just want to know that how can I do the follow cmd via python urllib2:
"curl -D- -u admin:admin -X POST -H "X-Atlassian-Token: nocheck" -F "file=#myfile.txt" /rest/api/2/issue/TEST-123/attachments"
And I don't want use subprocess.popen

The key to getting this to work was in setting up the multipart-encoded files:
import requests
# Setup authentication credentials
credentials = requests.auth.HTTPBasicAuth('USERNAME','PASSWORD')
# JIRA required header (as per documentation)
headers = { 'X-Atlassian-Token': 'no-check' }
# Setup multipart-encoded file
files = [ ('file', ('file.txt', open('/path/to/file.txt','rb'), 'text/plain')) ]
# (OPTIONAL) Multiple multipart-encoded files
files = [
('file', ('file.txt', open('/path/to/file.txt','rb'), 'text/plain')),
('file', ('picture.jpg', open('/path/to/picture.jpg', 'rb'), 'image/jpeg')),
('file', ('app.exe', open('/path/to/app.exe','rb'), 'application/octet-stream'))
]
# Please note that all entries are called 'file'.
# Also, you should always open your files in binary mode when using requests.
# Run request
r = requests.post(url, auth=credentials, files=files, headers=headers)
https://2.python-requests.org/en/master/user/advanced/#post-multiple-multipart-encoded-files

As in official documentation, we need to open the file in binary mode and then upload. I hope below small piece of code helps you :)
from jira import JIRA
# Server Authentication
username = "XXXXXX"
password = "XXXXXX"
jira = JIRA(options, basic_auth=(str(username), str(password)))
# Get instance of the ticket
issue = jira.issue('PROJ-1')
# Upload the file
with open('/some/path/attachment.txt', 'rb') as f:
jira.add_attachment(issue=issue, attachment=f)
https://jira.readthedocs.io/en/master/examples.html#attachments

You can use the jira-python package.
Install it like this:
pip install jira-python
To add attachments, use the add_attachment method of the jira.client.JIRA class:
add_attachment(*args, **kwargs) Attach an attachment to an issue and returns a Resource for it. The client will not attempt
to open or validate the attachment; it expects a file-like object to
be ready for its use. The user is still responsible for tidying up
(e.g., closing the file, killing the socket, etc.)
Parameters:
issue – the issue to attach the attachment
to attachment – file-like object to
attach to the issue, also works if it is a string with the
filename. filename – optional name for
the attached file. If omitted, the file object’s name attribute
is used. If you aquired the file-like object by any other method than
open(),
make sure that a name is specified in one way or the other.
You can find out more information and examples in the official documentation

Sorry for my unclear question
Thanks to How to POST attachment to JIRA using REST API?.
I have already resolve it.
boundary = '----------%s' % ''.join(random.sample('0123456789abcdef', 15))
parts = []
parts.append('--%s' % boundary)
parts.append('Content-Disposition: form-data; name="file"; filename="%s"' % fpath)
parts.append('Content-Type: %s' % 'text/plain')
parts.append('')
parts.append(open(fpath, 'r').read())
parts.append('--%s--' % boundary)
parts.append('')
body = '\r\n'.join(parts)
url = deepcopy(self.topurl)
url += "/rest/api/2/issue/%s/attachments" % str(jrIssueId)
req = urllib2.Request(url, body)
req.add_header("Content-Type", "multipart/form-data; boundary=%s" % boundary)
req.add_header("X-Atlassian-Token", "nocheck")
res = urllib2.urlopen(req)
print res.getcode()
assert res.getcode() in range(200,207), "Error to attachFile " + jrIssueId
return res.read()