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Python Flask Basics Welcome to Flask View Args Through URL

I added name="Treehouse" to the function argument, however I still get " '/' without a name gives non 200 responce"

from flask import Flask

app = Flask(name)

@app.route('/<name>') def hello(name="Treehouse"): return "Hello {}".format(name)

flask_app.py
from flask import Flask

app = Flask(__name__)


@app.route('/<name>')
def hello(name="Treehouse"):
     return "Hello {}".format(name)

2 Answers

Brandon Oakes
seal-mask
.a{fill-rule:evenodd;}techdegree
Brandon Oakes
Python Web Development Techdegree Student 11,501 Points

You are very close to passing. The final question wants you to have a default (name="Treehouse") in case no name is passed. You did that part correctly, strong work. So you have the default name for a reason right, its to catch a url that does DOES NOT have a name variable. So you need to add another decorator to also catch a url without the name variable provided. The code below worked for me.

from flask import Flask

app = Flask(name)

@app.route('/') @app.route('/<name>') def hello(name="Treehouse"): return "Hello {}".format(name)

Thanks! makes sense, and the code works perfect now