'Flask shows TypeError: send_from_directory() missing 1 required positional argument: 'path'

When I deploy my Flask app on Azure, the view raises TypeError: send_from_directory() missing 1 required positional argument: 'path'. This isn't happening when I run locally.

from flask import send_from_directory

@app.route('/download/<path:filename>', methods=['GET', 'POST'])
def download(filename):
    uploads = os.path.join(app.root_path, app.config['UPLOAD_FOLDER'])
    return send_from_directory(directory=uploads, filename=filename)

python flask

Solution 1:^[1]

Change the final line to return send_from_directory(uploads, filename).

See the Flask docs about send_from_directory. The changelog at the bottom that says "Changed in version 2.0: path replaces the filename parameter."

If you still want to use named parameters, change filename= to path=. send_from_directory(directory=uploads, path=filename)

Solution 2:^[2]

In my case I need certificate, it is saved inside static/pdf/certficates folder

@app.route('/download/<filename>', methods = ["GET", "POST"])
def download(filename):
   uploads = os.path.join(current_app.root_path, "static/pdf/folder_name")
   return send_from_directory(directory=uploads,path=filename,as_attachment=True)

Solution 3:^[3]

return send_from_directory(directory=uploads, filename=filename)

change to

return send_from_directory(directory=uploads, path=filename, as_attachment=True)

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	davidism
Solution 2	Ganesan J
Solution 3	Ihor Konovalenko

'Flask shows TypeError: send_from_directory() missing 1 required positional argument: 'path'

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

Sources

Related Questions

Solution 1:^[1]

Solution 2:^[2]

Solution 3:^[3]