Flask view raises TypeError got unexpected keyword argument
flask, python
Solution
You defined the route to be `/enviaplaca/<placa>`, but you defined the view function without the `placa` argument. The URL captures need to match the function arguments.
@app.route('/echoplaca/<placa>')
def echoplaca(placa):
Problem
I'm trying to make a request to the following view. However, I get the error `TypeError: echoplaca() got an unexpected keyword argument 'placa'`. ``` @app.route(r'/enviaplaca/<placa>') def echoplaca(): return "Numero de placa: {}".format(placa) ```