Redirecting to URL in Flask

Question:

I’m trying to do the equivalent of Response.redirect as in C# – i.e.: 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)
Asked By: iJade

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Answers:

From the Flask API Documentation (v. 2.0.x):

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.
Answered By: ford
flask.redirect(location, code=302)

Docs can be found here.

Answered By: der_fenix

You have to return a redirect:

import os
from flask import Flask,redirect

app = Flask(__name__)

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

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)

See the documentation on flask docs. The default value for code is 302 so code=302 can be omitted or replaced by other redirect code (one in 301, 302, 303, 305, and 307).

Answered By: Xavier Combelle
#!/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.

Answered By: soerface

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.

Answered By: ivanleoncz

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.

Answered By: Manan Gouhari

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.

Answered By: alfredo

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.

Answered By: RAJAHMAD MULANI

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

Answered By: husseinraed

How to Redirect Users / Requests in Flask

Throwing an Error inside of your API handler function will redirect your user to an error handler, which can handle redirection. Alternatively you can just call redirect like everyone else is saying, but this is another way of redirecting unauthorized users. To demonstrate what I mean, I’ve provided an example below.

In a case where Users should be Authorized

First lets assume you have a protected route of which you protected like this.

def handle_api_auth(func):
    """
    **handle_api_auth**
        wrapper to handle public api calls authentications

    :param func: a function to be wrapped
    :return: wrapped function
    """

    @functools.wraps(func)
    def auth_wrapper(*args, **kwargs):
        api_key: Optional[str] = request.headers.get('x-api-key')
        secret_token: Optional[str] = request.headers.get('x-secret-token')
        domain: Optional[str] = request.base_url
        if is_request_valid(api_key=api_key, secret=secret_token, domain=domain):
            return func(*args, **kwargs)
        # NOTE: throwing an Error Here will redirect your user to an error handler or alteratively you can just call redirect like everyone else is saying, but this is another way of redirecting unathorized users
        message: str = "request not authorized"
        raise UnAuthenticatedError(status=error_codes.un_auth_error_code, description=message)

    return auth_wrapper

Definition of is_request_valid is as follows

@app_cache.cache.memoize(timeout=15 * 60, cache_none=False)  # timeout equals fifteen minutes // 900 seconds
def is_request_valid(api_key: str, secret: str, domain: str) -> bool:
    """
    **is_api_key_valid**
        validates api keys on behalf of client api calls

    :param api_key: str -> api_key to check
    :param secret: str -> secret token
    :param domain: str -> domain registered for the api_key and secret_token
    :return: bool -> True if api_key is valid
    """

    organization_id: str = config_instance.ORGANIZATION_ID
    # NOTE: lets assumy api_keys_view.get_api_key will return the api keys from some database somewhere
    response = api_keys_view.get_api_key(api_key=api_key, organization_id=organization_id)

    response_data, status_code = response
    response_dict = response_data.get_json()

    if not response_dict.get('status'):
        return False

    api_instance: dict = response_dict.get('payload')
    if not isinstance(api_instance, dict):
        return False

    domain: str = domain.lower().strip()
    # NOTE accessing the keys this way will throw ValueError if keys are not available which is what we want
    # Any Error which gets thrown Ridirects the Users from the path the user is on to an error handler.
    is_secret_valid: bool = hmac.compare_digest(api_instance['secret_token'], secret)
    is_domain_valid: bool = hmac.compare_digest(api_instance['domain'], domain)
    _request_valid: bool = is_secret_valid and is_domain_valid

    return not not api_instance.get('is_active') if _request_valid else False

Define your Error Handlers like this

from flask import Blueprint, jsonify, request, redirect
from werkzeug.exceptions Unauthorized

error_handler = BluePrint('error_handlers', __name__)

@error_handler.app_errorhandler(Unauthorized)
def handle_error(e : Unauthorized) -> tuple:
    """default unath handler"""
    return jsonify(dict(message=e.description)), e.code if request.headers.get('content-type') == 'application/json' else redirect('/login')

handle other errors the same and note that in-case the request was

not a json the user gets redirected to a login page
if json the user gets sent an unathecated response then its
up to the front end to handle Unath Errors..

Answered By: mobius-crypt

There are two ways you can redirect to a URL in Flask.

  1. You want to for example, redirect a user to another route after he or she login, etc.
  2. You might also want to redirect a user to a route that expect some variable example: @app.route('/post/<string:post_id>')

Well, to implement flask redirect for case # 1, its simple, just do:

from flask import Flask,redirect,render_template,url_for
app = Flask(__name__)


@app.route('/login')
def login():
    # if user credentials are valid, redirect to user dashboard
    if login == True:
       return redirect(url_for(app.dashboard))
    else:
       print("Login failed !, invalid credentials")
    return render_template('login.html',title="Home Page")


@app.route('/dashboard')
def dashboard():
    return render_template('dashboard.html',title="Dashboard")

To implement flask redirect for case #2, do the following

from flask import Flask,redirect,render_template,url_for
app = Flask(__name__)


@app.route('/home')
def home():
    # do some logic, example get post id
    if my_post_id:
       # **Note:** post_id is the variable name in the open_post route
       # We need to pass it as **post_id=my_post_id**
       return redirect(url_for(app.open_post,post_id=my_post_id))
    else:
       print("Post you are looking for does not exist")
    return render_template('index.html',title="Home Page")


@app.route('/post/<string:post_id>')
def open_post():
    return render_template('readPost.html',title="Read Post")

Same thing can be done in view

<a href="{{url_for(app.open_post,post_id=my_post_id)}}"></a>

Please Note: when redirecting always use the app.home or app.something.. (route or view function name) instead of using redirect("/home").
Reason is, if you modify the route example from "/home" to "/index/page" for some reason, then your code will break

Answered By: Randall Moore
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