AssertionError: View function mapping is overwriting an existing endpoint function: main

Your view names need to be unique even if they are pointing to the same view method.

app.add_url_rule('/',
view_func=Main.as_view('main'),
methods = ['GET'])


app.add_url_rule('/<page>/',
view_func=Main.as_view('page'),
methods = ['GET'])

There is a fix for Flask issue #570 introduced recenty (flask 0.10) that causes this exception to be raised.

See https://github.com/mitsuhiko/flask/issues/796

So if you go to flask/app.py and comment out the 4 lines 948..951, this may help until the issue is resovled fully in a new version.

The diff of that change is here: http://github.com/mitsuhiko/flask/commit/661ee54bc2bc1ea0763ac9c226f8e14bb0beb5b1

Flask requires you to associate a single 'view function' with an 'endpoint'. You are calling Main.as_view('main') twice which creates two different functions (exactly the same functionality but different in memory signature). Short story, you should simply do

main_view_func = Main.as_view('main')


app.add_url_rule('/',
view_func=main_view_func,
methods=["GET"])


app.add_url_rule('/<page>/',
view_func=main_view_func,
methods=["GET"])

This same issue happened to me when I had more than one API function in the module and tried to wrap each function with 2 decorators:

1. @app.route()
2. My custom @exception_handler decorator

I got this same exception because I tried to wrap more than one function with those two decorators:

@app.route("/path1")
@exception_handler
def func1():
pass


@app.route("/path2")
@exception_handler
def func2():
pass

Specifically, it is caused by trying to register a few functions with the name wrapper:

def exception_handler(func):
def wrapper(*args, **kwargs):
try:
return func(*args, **kwargs)
except Exception as e:
error_code = getattr(e, "code", 500)
logger.exception("Service exception: %s", e)
r = dict_to_json({"message": e.message, "matches": e.message, "error_code": error_code})
return Response(r, status=error_code, mimetype='application/json')
return wrapper

Changing the name of the function solved it for me (wrapper.__name__ = func.__name__):

def exception_handler(func):
def wrapper(*args, **kwargs):
try:
return func(*args, **kwargs)
except Exception as e:
error_code = getattr(e, "code", 500)
logger.exception("Service exception: %s", e)
r = dict_to_json({"message": e.message, "matches": e.message, "error_code": error_code})
return Response(r, status=error_code, mimetype='application/json')
# Renaming the function name:
wrapper.__name__ = func.__name__
return wrapper

Then, decorating more than one endpoint worked.

For users that use @app.route it is better to use the key-argument endpoint rather then chaning the value of __name__ like Roei Bahumi stated. Taking his example will be:

@app.route("/path1", endpoint='func1')
@exception_handler
def func1():
pass


@app.route("/path2", endpoint='func2')
@exception_handler
def func2():
pass

If you think you have unique endpoint names and still this error is given then probably you are facing issue. Same was the case with me.

This issue is with flask 0.10 in case you have same version then do following to get rid of this:

sudo pip uninstall flask
sudo pip install flask=0.9

I would just like to add to this a more 'template' type solution.

def func_name(f):
def wrap(*args, **kwargs):
if condition:
pass
else:
whatever you want
return f(*args, **kwargs)
wrap.__name__ = f.__name__
return wrap

would just like to add a really interesting article "Demystifying Decorators" I found recently: https://sumit-ghosh.com/articles/demystifying-decorators-python/

Your view names need to be unique even if they are pointing to the same view method, or you can add from functools import wraps and use @wraps https://docs.python.org/2/library/functools.html

use flask 0.9 instead use the following commands sudo pip uninstall flask

sudo pip install flask==0.9

This can happen also when you have identical function names on different routes.

Adding @wraps(f) above the wrapper function solved my issue.

def list_ownership(f):
@wraps(f)
def decorator(*args,**kwargs):
return f(args,kwargs)
return decorator

In case you are using flask on python notebook, you need to restart kernel everytime you make changes in code

This is issue for me was from an (breaking) update to flask-jwt-extended (version 4.x.x and up) used in a basic api I wrote a year ago and am now incorporating into a project.

@jwt_required to @jwt_required()

maybe something contains no difference

1. check url
2. check function name

I'm working on a similar problem and I managed to get rid of it by returning the wrapper funcion, which wasn't being done before:

def decorator_func(func_to_decorate):
def wrapper_func():
return func_to_decorate
return wrapper_func     # I wasn't returning wrapper func!

I encountered the same AssertionError, but have not seen a solution here. In my case, I run a decorated function in a Jupyter cell twice, then the error was caused by running an endpoint function with the same name as an existing running endpoint function.