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)
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).
#!/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.
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.
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.
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.
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.
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')
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)
