How to upload file with python requests?

ID : 20119

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Tags : pythonfilefile-uploadpython-requestspython

Top 5 Answer for How to upload file with python requests?

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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 =, 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', '', 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.

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(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 = '' files = {'file': open('report.xls', 'rb')}  r =, files=files) r.text 
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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:'http://some.url/streamed', data=f) 

Server Side

Then store the file on the 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(['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')"start receiving file ... filename => " + str(         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():" ".join(["saved form name",, "submitted as", fil.filename, "to temporary file",]))         # Do whatever with stored file at ``     return 'OK', 200 
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@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 =, 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:         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 =, 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.

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In Ubuntu you can apply this way,

to save file at some location (temporary) and then open and send it to API

      path ='static/tmp/' +, ContentFile(       path12 = os.path.join(os.getcwd(), "static/tmp/" +       data={} #can be anything u want to pass along with File       file1 = open(path12, 'rb')       header = {"Content-Disposition": "attachment; filename=" +, "Authorization": "JWT " + token}        res=,data,header) 

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