Flask upload: How to get file name?

28,580

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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Amrit Gill
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Amrit Gill

Updated on July 09, 2022

Comments

  • Amrit Gill
    Amrit Gill almost 2 years

    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?

  • Rivenfall
    Rivenfall about 4 years
    why does file.filename give me the URL path instead of the file name :-/ ?