Does anyone know if flask redirect is able to open a link / URL in a new tab?
#app.route('/test')
def my_page():
return redirect('http://mylink.com', 301)
I would like if the mylink website opened in a new tab, any insights?
As far as I know that would not be a flask issue. You have to open a new tab within your html code or with Javascript.
example:
Link
The server has no power over what the browser does in this case.
You could also use Python's webbrowser module to open a page, if you'd like to avoid getting into the HTML
import webbrowser
#app.route('/test')
def my_page():
return webbrowser.open_new_tab('http://mylink.com')
If you're dealing with a form you can set target="_blank"
A slight adjustment to #garettmd that fixes #John Jiang's issue:
import webbrowser
from flask import Flask, redirect, url_for
#app.route("/test/")
def test():
webbrowser.open("https://google.com")
return redirect(url_for("index"))
This way we get to return a redirect (which we should probably do anyway) and we also get our new tab without "The view function did not return a valid response" error.
Related
So I have been trying to get used to Flash in python but I've come across a problem. I want that when http://localhost:5000/ is inserted in the browser a html page is displayed. I've tried multiple ways to do this like using the render_template() but that returns me a jinja2.exceptions.TemplateNotFound: index.html. I've also tried a simple return redirect() but that throws something saying the adress was not recognized or understood. When I tried using the url_for() it threw 404 - not found. I really have no idea how to fix this.
# htppserver.py
import flask
import threading
app = Flask(__name__, template_folder="Dashboard/website")
#app.route("/site", methods=["GET"])
#app.route("/", methods=["GET"])
def get_site():
return render_template("index.html")
x = threading.Thread(target=app.run)
x.start()
Currently my dir system looks something like this
main_folder # This is the working directory accordingly to os.getcwd()
├──cogs
│ └──httpserver.py # Source code is here
└──Dashboard
└website
├──...
├──index.html # This is the file I want to show
└──...
Thanks
put your html files in a folder called "templates" in the same directory as the python file that serves
I want to redirect to external link without quiting current domain link.
Following is an example of my question :
https://translate.google.com/translate?
depth=1&hl=ar&prev=search&rurl=translate.google.com&sl=en&sp=nmt4&u=https://bookboon.com/
As you can see the main website here is google translate and second is bookbone.com
How can we do that with python using a framework like Django2.0?
Thank you.
Put the below code in views.py and make URL for that
import urlparse
from django.http import HttpResponseRedirect
domain = request.GET['domain']
destination = reverse('variable_response',args=['Successful'])
full_address = urlparse.urljoin(domain, destination)
return HttpResponseRedirect(full_address)
I would like to know how I can modify the URL to the welcome page.
Currently it is /superset/welcome.
It is run into superset/views/core.py in a #expose('/welcome').
I know I can modify the code inside this #expose, but I want to redirect to another url.
So I want to find the line where there is:
welcome_page = /superset/welcome
As of Superset 1.3, you can change the default landing page by adding this code to your Superset config:
from flask import Flask, redirect
from flask_appbuilder import expose, IndexView
from superset.typing import FlaskResponse
class SupersetDashboardIndexView(IndexView):
#expose("/")
def index(self) -> FlaskResponse:
return redirect("/dashboard/list/")
FAB_INDEX_VIEW = f"{SupersetDashboardIndexView.__module__}.{SupersetDashboardIndexView.__name__}"
In the above example, I am using /dashboard/list/ instead of the default /superset/welcome/.
The code above is Unlicensed and thus is free and unencumbered software released into the public domain.
In superset's file structure, navigate to:
superset/app.py
There you will find
class SupersetIndexView(IndexView):
#expose("/")
def index(self) -> FlaskResponse:
return redirect("/superset/welcome")
Modify this to path where you want to redirect.
I want to put a button on a flask-generated web page and let a user download the html page as a file when the user clicks the button. What I imagine is something like saving the rendered html into BytesIO and send it via send_file, but I can't find how to save the rendered page into a file object. How can I do that?
You could try something like this:
import StringIO
from flask import Flask, send_file, render_template
def page_code():
strIO = StringIO.StringIO()
strIO.write(render_template('hello.html', name='World'))
strIO.seek(0)
return send_file(strIO,
attachment_filename="testing.txt",
as_attachment=True)
It is not tested but should give you an idea.
#gtomer Using the input from the thread, I may have figured out a solution that helped eliminate the AssertionError (by using BytesIO since on StringIO) that worked for me. This is on Python39.
xx = asteriod.query.filter_by(order_number=tagVV).first()
yy = toposphere.query.filter_by(order_number=tagVV).all()
apple = io.BytesIO()
apple.write(render_template("receipt2.html", x12=yy, x13=ww, x14=xx).encode('utf-8'))
apple.seek(0)
return send_file(apple, attachment_filename="testing.html", as_attachment=True)
When I run this code on my computer with the help of "Google App Engine SDK", it displays (in my browser) the HTML code of the Google home page:
from google.appengine.api import urlfetch
url = "http://www.google.com/"
result = urlfetch.fetch(url)
print result.content
How can I make it display the page itself? I mean I want to see that page in my browser the way it would normally be seen by any user of the internet.
Update 1:
I see I have received a few questions that look a bit complicated to me, although I definitely remember I was able to do it, and it was very simple, except i don't remember what exactly i changed then in this code.
Perhaps, I didn't give You all enough details on how I run this code and where I found it. So, let me tell You what I did. I only installed Python 2.5 on my computer and then downloaded "Google App Engine SDK" and installed it, too. Following the instructions on "GAE" page (http://code.google.com/appengine/docs/python/gettingstarted/helloworld.html) I created a directory and named it “My_test”, then I created a “my_test.py” in it containing that small piece of the code that I mentioned in my question.
Then, continuing to follow on the said instructions, I created an “app.yaml” file in it, in which my “my_test.py” file was mentioned. After that in “Google App Engine Launcher” I found “My_test” directory and clicked on Run button, and then on Browse. Then, having visited this URL http://localhost:8080/ in my web browser, I saw the results.
I definitely remember I was able to display any page in my browser in this way, and it was very simple, except I don’t remember what exactly I changed in the code (it was a slight change). Now, all I can see is a raw HTML code of a page, but not a page itself.
Update 2:
(this update is my response to wescpy)
Hello, wescpy!!! I've tried Your updated code and something didn't work well there. Perhaps, it's because I am not using a certain framework that I am supposed to use for this code. Please, take a look at this screen shot (I guess You'll need to right-click this image to see it in better resolution):
(source: narod.ru)
Is not that easy, you have to parse content and adjust relative to absolute paths for images and javascripts.
Anyway, give it a try adding the correct Content-Type:
from google.appengine.api import urlfetch
url = "http://www.google.com/"
result = urlfetch.fetch(url)
print 'Content-Type: text/html'
print ''
print result.content
a more complete example would look something like this:
from google.appengine.ext import webapp
from google.appengine.ext.webapp.util import run_wsgi_app
from google.appengine.api import urlfetch
class MainHandler(webapp.RequestHandler):
def get(self):
url = "http://www.google.com/"
result = urlfetch.fetch(url)
self.response.out.write(result.content)
application = webapp.WSGIApplication([
('/', MainHandler),
], debug=True)
def main():
run_wsgi_app(application)
if __name__ == '__main__':
main()
but as others' have said, it's not that easy to do because you're not in the server's domain, meaning the pages will likely not look correct due to missing static content (JS, CSS, and/or images)... unless full pathnames are used or everything that's needed is embedded into the page itself.
UPDATE 1:
as mentioned before, you cannot just download the HTML source and expect things to render correctly because you don't necessarily have access to the static data. if you really want to render it as it was meant to be seen, you have to just redirect... here's the modified piece of code:
from google.appengine.ext import webapp
from google.appengine.ext.webapp.util import run_wsgi_app
from google.appengine.api import urlfetch
class MainHandler(webapp.RequestHandler):
def get(self):
url = "http://www.google.com/"
self.redirect(url)
application = webapp.WSGIApplication([
('/', MainHandler),
], debug=True)
def main():
run_wsgi_app(application)
if __name__ == '__main__':
main()
UPDATE 2:
sorry! it was a cut-n-paste error. now try it.
special characters such as <> etc are likely encoded, you'd have to decode them again for the browser to interpet it as code.