Question

Using app.open_resource('foobar.txt', 'w') generates the error Resources can only be opened for reading in flask.

Is there a way to open a resource to write to it?

If not, can you get the path of the resource using flask and then I can open it manually and write to it.

Answer 1

This should work:

import os
f = open(os.path.join(app.root_path, 'foobar.txt'), 'w')

This is more convenient:

import os
with open(os.path.join(app.root_path, 'foobar.txt'), 'w') as f:
    ...