[python] Redirecting to URL in Flask

I'm new to Python and Flask and I'm trying to do the equivalent of Response.redirect as in C# - ie: redirect to a specific URL - how do I go about this?

Here is my code:

import os
from flask import Flask

app = Flask(__name__)

@app.route('/')
def hello():
    return 'Hello World!'

if __name__ == '__main__':
    # Bind to PORT if defined, otherwise default to 5000.
    port = int(os.environ.get('PORT', 5000))
    app.run(host='0.0.0.0', port=port)

I believe that this question deserves an updated. Just compare with other approaches.

Here's how you do redirection (3xx) from one url to another in Flask (0.12.2):

#!/usr/bin/env python

from flask import Flask, redirect

app = Flask(__name__)

@app.route("/")
def index():
    return redirect('/you_were_redirected')

@app.route("/you_were_redirected")
def redirected():
    return "You were redirected. Congrats :)!"

if __name__ == "__main__":
    app.run(host="0.0.0.0",port=8000,debug=True)

For other official references, here.

For this you can simply use the redirect function that is included in flask

from flask import Flask, redirect

app = Flask(__name__)

@app.route('/')
def hello():
    return redirect("https://www.exampleURL.com", code = 302)

if __name__ == "__main__":
    app.run()

Another useful tip(as you're new to flask), is to add app.debug = True after initializing the flask object as the debugger output helps a lot while figuring out what's wrong.

Flask includes the redirect function for redirecting to any url. Futhermore, you can abort a request early with an error code with abort:

from flask import abort, Flask, redirect, url_for

app = Flask(__name__)

@app.route('/')
def hello():
    return redirect(url_for('hello'))

@app.route('/hello'):
def world:
    abort(401)

By default a black and white error page is shown for each error code.

The redirect method takes by default the code 302. A list for http status codes here.

its pretty easy if u just want to redirect to a url without any status codes or anything like that u can simple say

from flask import Flask, redirect

app = Flask(__name__)

@app.route('/')
def redirect_to_link():
    # return redirect method, NOTE: replace google.com with the link u want
    return redirect('https://google.com')

here is the link to the Flask Docs for more explanation

#!/usr/bin/env python
# -*- coding: utf-8 -*-

import os
from flask import Flask, redirect, url_for

app = Flask(__name__)

@app.route('/')
def hello():
    return redirect(url_for('foo'))

@app.route('/foo')
def foo():
    return 'Hello Foo!'

if __name__ == '__main__':
    # Bind to PORT if defined, otherwise default to 5000.
    port = int(os.environ.get('PORT', 5000))
    app.run(host='0.0.0.0', port=port)

Take a look at the example in the documentation.

You can use like this:

import os
from flask import Flask

app = Flask(__name__)

@app.route('/')
def hello():
     # Redirect from here, replace your custom site url "www.google.com"
    return redirect("https://www.google.com", code=200)

if __name__ == '__main__':
    # Bind to PORT if defined, otherwise default to 5000.
    port = int(os.environ.get('PORT', 5000))
    app.run(host='0.0.0.0', port=port)

Here is the referenced link to this code.

flask.redirect(location, code=302)

Docs can be found here.

From the Flask API Documentation (v. 0.10):

flask.redirect(location, code=302, Response=None)

Returns a response object (a WSGI application) that, if called, redirects the client to the target location. Supported codes are 301, 302, 303, 305, and 307. 300 is not supported because it’s not a real redirect and 304 because it’s the answer for a request with a request with defined If-Modified-Since headers.

New in version 0.6: The location can now be a unicode string that is encoded using the iri_to_uri() function.

Parameters:

• location – the location the response should redirect to.
• code – the redirect status code. defaults to 302.
• Response (class) – a Response class to use when instantiating a response. The default is werkzeug.wrappers.Response if unspecified.