How to redefine url_for in flask by automatically adding a string '/foo' as a prefix of the url generated from url_for?

Question

It may look very odd, but this is what I want to achieve exactly, it is not related with blueprints or add prefix to all routers. thanks!

Can you add examples of the desired output? – arshovon Aug 14 '19 at 06:24 — arshovon, Aug 14 '19 at 06:24

score 1 · Accepted Answer · answered Aug 14 '19 at 08:53

You can define your own:

# url_for_custom.py
import flask

def url_for(endpoint, **values):
    url = flask.url_for(endpoint, **values)
    if not values.get('_external', False):
        url = '/foo' + url
    return url

If you wish to change it globally, you can monkey-patch some time early on:

# __init__.py
import flask
from url_for_custom import url_for

flask.url_for = url_for

Although it's perhaps better to simply import and use your own implementation.

How to redefine url_for in flask by automatically adding a string '/foo' as a prefix of the url generated from url_for?

1 Answers1