flask route with default argument at beginning of path - flask

Question

In flask, I can specify a default argument to a route thus:
#app.route("/user", defaults={'uid': 3})
#user.route('/user/<uid>')
def user(uid):
# If /user or /user/ is requested, I'll see uid=3.
# If /user/4 is requested, I'll see uid=4.
But I have a reverse path:
#app.route("/user", defaults={'thing': None})
#user.route('/<thing>/user/')
def user(thing):
# If /user or /user/ is requested, I want to see thing == None.
# If /dog/user/ is requested, I'll see thing == 'dog'.
Basically, if thing isn't provided, I'll do something a bit complex to compute it. But it's fixed for a session, so I pass it around.
I know that this is asking quite a lot from the route parser. (In particular, no thing had better conflict with another path, and I could easily cause quite a lot of inefficient backtracking if I'm not careful. But things are designed never to conflict.)
The question is whether there's a way to do this?
What I've done is to create two functions. But I have a handful of functions like this, so this is feeling heavy.
#app.route("/user")
def user():
return redirect(url_for('main.user', thing=ComputeThing()))
#user.route('/<thing>/user/')
def user(thing):
# If /user or /user/ is requested, I want to see thing == None.
# If /dog/user/ is requested, I'll see thing == 'dog'.

Answer 1

I must be missing something, can you expand on what you mean by a reverse path, because just using defaults seems to achieve what you want in your last example.
from flask import Flask
from random import choice
app = Flask(__name__)
choices = ['fox', 'cat', 'eel', 'cow']
def compute_thing():
return choice(choices)
#app.route("/user", defaults={'thing': None})
#app.route("/<thing>/user/")
def user(thing):
if thing is None:
thing = compute_thing()
return f"`thing` set as {thing}"
if __name__ == '__main__':
tc = app.test_client()
ru = tc.get('/user')
print(ru.data) # `thing` set as cow'
rd = tc.get('/dog/user/')
print(rd.data) # `thing` set as dog'