how to get static files in Flask without url_for('static', file_name='xxx')

Question

I don't want use url_for('static', file_name='foo.jpg') to get static file in template.

how to get static file in this way:

<img src="/pic/foo.jpg" />

thanks

As long as `/pic/foo.jpg` exists it should work fine. Did you try it? — Cfreak, Dec 04 '12 at 15:35
@Cfreak thank you , it's works fine. i write the wrong path before. — Robin, Dec 05 '12 at 03:17
@IgnasB. i think common path more convenient and intuitive for FE, what's your point — Robin, Dec 05 '12 at 03:26
@Robin, well just asked :) I thought maybe you spotted some performance issue or something. I always prefer to have dynamic things, just in case I will change static dir to another or smth, so I don't ned to worry then about the some time ago created template which I already forgot to exist :) But this is everyone's choice ;) — Ignas Butėnas, Dec 05 '12 at 06:20
i've just had the same problem and solved it like that: http://stackoverflow.com/a/29521067/303114 — danfromisrael, Apr 09 '15 at 09:46
i just had a similar issue and i solved like that: http://stackoverflow.com/a/29521067/303114 notice it's quite similar to what i_4_got is suggesting just using add_url_rule method instead of the route — danfromisrael, Apr 09 '15 at 09:52

score 17 · Accepted Answer · edited Nov 04 '14 at 08:01

You can set up your own route to serve static files. Add this method and update the static path directory in the send_from_directory method, then your img tag should work.

@app.route('/pic/<path:filename>')
def send_pic(filename):
    return send_from_directory('/path/to/static/files', filename)

For a production app, you should set up your server to serve static files directly. It would be much faster and use less server resources, but for a few users the difference shouldn't be a problem.

how to get static files in Flask without url_for('static', file_name='xxx')

1 Answers1