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A client is sending files with arbitrary names. I am handling the request with the following implementation.
@app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
# 'file-name' is the file name here
if 'file-name' not in request.files:
flash('No file part')
return 'no file found'
file = request.files['file-name']
Should I ask another query/path parameter which defines the file name?
611
Once you fetch the actual file with file = request.files['file'], you can get the filename with file.filename.
The documentation provides the following complete example. (Note that 'file' is not the filename; it's the name of the form field containing the file.)
def allowed_file(filename):
return '.' in filename and \
filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
# check if the post request has the file part
if 'file' not in request.files:
flash('No file part')
return redirect(request.url)
file = request.files['file']
# if user does not select file, browser also
# submit a empty part without filename
if file.filename == '':
flash('No selected file')
return redirect(request.url)
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return redirect(url_for('uploaded_file',
filename=filename))
return '''
<!doctype html>
<title>Upload new File</title>
<h1>Upload new File</h1>
<form method=post enctype=multipart/form-data>
<p><input type=file name=file>
<input type=submit value=Upload>
</form>
'''
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