python requests upload file

Question

I'm visiting a website and I want to upload a file.

I wrote the code in python:

import requests

url = 'http://example.com'
files = {'file': open('1.jpg', 'rb')}
r = requests.post(url, files=files)
print(r.content)

But it seems no file has been uploaded, and the page is the same as the initial one.

I wonder how I could upload a file.

The source code of that page:

<html><head><meta charset="utf-8" /></head>

<body>
<br><br>
Upload<br><br>
<form action="upload.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="hidden" name="dir" value="/uploads/" />
<input type="file" name="file" id="file" /> 
<br />
<input type="submit" name="submit" value="Submit" />
</form>

</body>
</html>

Btw, you are not sending dir. r = requests.post(url, files=files, data={"dir": "/uploads/"}) — Ozgur Vatansever
– Ozgur Vatansever, Commented May 12, 2017 at 13:21
@OzgurVatansever I have added the data. But still no file is uploaded. — Arolia
– Arolia, Commented May 12, 2017 at 13:27

coffee-grinder · Accepted Answer · 2018-01-11 00:44:01Z

14

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)

edited Jan 11, 2018 at 0:44

coffee-grinder

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answered May 12, 2017 at 16:39

t.m.adam

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3 Comments

Arolia Over a year ago

Thanks. It works now. It seems that I have the wrong url to upload the file.

stevezkw Over a year ago

Without passing the param 'headers', it will raise an error : werkzeug.exceptions.UnsupportedMediaType: 415 Unsupported Media Type: Did not attempt to load JSON data because the request Content-Type was not 'application/json'.

t.m.adam Over a year ago

@stevezkw Yes, I think the Content-Type header is set to multipart/form-data automatically by requests. Which is what we want when uploading a file. But in your case you should probably use the json= parameter

Ahmed Ginani · Accepted Answer · 2017-05-12 13:35:33Z

3

First of all, define path of upload directory like,

app.config['UPLOAD_FOLDER'] = 'uploads/'

Then define file extension which allowed to upload like,

app.config['ALLOWED_EXTENSIONS'] = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'gif'])

Now suppose you call function to process upload file then you have to write code something like this,

# Route that will process the file upload
@app.route('/upload', methods=['POST'])
def upload():
    # Get the name of the uploaded file
    file = request.files['file']

    # Check if the file is one of the allowed types/extensions
    if file and allowed_file(file.filename):
        # Make the filename safe, remove unsupported chars
        filename = secure_filename(file.filename)

        # Move the file form the temporal folder to
        # the upload folder we setup
        file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))

        # Redirect the user to the uploaded_file route, which
        # will basicaly show on the browser the uploaded file
        return redirect(url_for('YOUR REDIRECT FUNCTION NAME',filename=filename))

This way you can upload your file and store it in your located folder.

I hope this will help you.

Thanks.

answered May 12, 2017 at 13:35

Ahmed Ginani

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1 Comment

Arolia Over a year ago

Thanks for your reply. Actually, I am just visiting the website owned by others. I can upload the file through my browser, but I need to find a way to upload it through python.

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