intelis intelis - 4 months ago 86
Python Question

Werkzeug(Flask) response with redirect not working

I'm building a simple

Flask
application and I want to return redirect response. Also, I want to maintain total control over headers.

This is what I've got so far:

from flask import Flask
from werkzeug.wrappers import Response

app = Flask(__name__)

@app.route('/toexample')
def to_example():

headers = {
'User-Agent': 'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:39.0) Gecko/20100101 Firefox/39.0',
'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
'Accept-Language': 'en-US,en;q=0.5',
'Accept-Encoding': 'gzip, deflate',
}

return Response('www.example.com', status_code=302, headers=headers)

if __name__ == '__main__':
app.run(debug=True)


I get an error:

TypeError: __init__() got an unexpected keyword argument 'status_code'


Ok, it looks like
status_code
doesn't exist on
__init__()
, but what is the right way of doing this?

What I ultimately want is a link that a user would click but again, I want to maintain control over headers(Connection, Cookies, Referer, etc.)

Answer

I had to add Location to headers. Also, status_code is wrong, it should've been status=302.

Working example:

from flask import Flask
from werkzeug.wrappers import Response

app = Flask(__name__)

@app.route('/toexample')
def to_example():

    headers = {
            'User-Agent': 'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:39.0) Gecko/20100101 Firefox/39.0',
            'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
            'Accept-Language': 'en-US,en;q=0.5',
            'Accept-Encoding': 'gzip, deflate',
            'Location': 'http://www.example.com'
    }

    return Response('http://www.example.com', status=302, headers=headers)

if __name__ == '__main__':
    app.run(debug=True)