My question is, can an endpoint in a Flask API call a method from another endpoint, that is a member of itself? Is there a proper way of doing this?
Yes, you can indeed call another endpoint by using redirect with url_for
If this is something you are trying to achieve.
from flask import Flask, redirect, url_for
app = Flask(__name__)
#app.route('/')
def hello_world():
return "Hello world"
#app.route('/home')
def hello_again():
return redirect(url_for('.hello_world'))
if __name__=="__main__":
app.run(debug=True)
The request to /home will redirect you to /.
Related
This question already has an answer here:
Redirect back in Flask
(1 answer)
Closed 2 years ago.
I would like to use flask to define the action for a button. However, after triggering the action, I will automatically be redirected. E.g. having a button definition using the following html:
<html>
<body>
<form action="/do_something" method="post">
<button>Do something!</button>
</form>
</body>
</html>
And the action handling by the given python script:
from flask import Flask, render_template
app = Flask(__name__)
#app.route("/")
def root():
return render_template("test_3.html")
#app.route("/do_something", methods=["POST"])
def something():
return 'Something'
app.run()
In this case, I start with the url "/". By clicking on the button "Do Something", I will automatically be redirected to the url "/do_something". What I actual want is:
Perform Action
Refresh the original page
Can some one help?
If the POST request to /do_something can occur on any page, I would recommend redirecting back to the request's referrer, which will "refresh" the page.
E.g
from flask import Flask, render_template, request, redirect
app = Flask(__name__)
#app.route("/")
def root():
return render_template("test_3.html")
#app.route("/do_something", methods=["POST"])
def something():
# Do things
return redirect(request.referrer or request.host_url)
app.run()
Otherwise you can specify the page to go to with redirect() and url_for()
Example:
from flask import Flask, redirect, render_template, url_for
app = Flask(__name__)
#app.route("/")
def root():
return render_template("test_3.html")
#app.route("/do_something", methods=["POST"])
def something():
# Do things
return redirect(url_for("app.root"))
app.run()
I inherited a project that uses flask. This flask application has several APIs, each API has a GET function that returns a json object.
I was asked to implement an additional API that requests information from the other APIs. So my question is how do I make a GET request directly to a flask application? Is it something like....
from flask import request
#app.route('/root_dir/api_number_1/info', methods=['GET'])
def request_info_from_api_number_1():
return request.get_json()
#app.route('/root_dir/api_number_2/info', methods=['GET'])
def request_info_from_api_number_2():
return request.get_json()
When I do this the functions return None. I suppose I could always make an http request to the flask url and specify the address as localhost but it seems strange to have to make an http request when I can directly access the flask app object.
Could you just use the existing api functions?
from flask import Flask
app = Flask(__name__)
#app.route("/foo")
def foo():
return "foo"
#app.route("/foobar")
def foobar():
return foo() + "bar"
if __name__ == "__main__":
app.run(debug=True)
I'm building the Flask app with React, I ended up having a problem with routing.
The backend is responsible to be an API, hence some routes look like:
#app.route('/api/v1/do-something/', methods=["GET"])
def do_something():
return something()
and the main route which leads to the React:
#app.route('/')
def index():
return render_template('index.html')
I'm using react-router in the React app, everything works fine, react-router takes me to /something and I get the rendered view, but when I refresh the page on /something then Flask app takes care of this call and I get Not Found error.
What is the best solution? I was thinking about redirecting all calls which are not calling /api/v1/... to / it's not ideal as I will get back the home page of my app, not rendered React view.
We used catch-all URLs for this.
from flask import Flask
app = Flask(__name__)
#app.route('/', defaults={'path': ''})
#app.route('/<path:path>')
def catch_all(path):
return 'You want path: %s' % path
if __name__ == '__main__':
app.run()
You can also go an extra mile and reuse the Flask routing system to match path to the same routes as client so you can embed the data client will need as JSON inside the HTML response.
Maybe as extension to the answers before. This solved the problem for me:
from flask import send_from_directory
#app.route('/', defaults={'path': ''})
#app.route('/<path:path>')
def serve(path):
path_dir = os.path.abspath("../build") #path react build
if path != "" and os.path.exists(os.path.join(path_dir, path)):
return send_from_directory(os.path.join(path_dir), path)
else:
return send_from_directory(os.path.join(path_dir),'index.html')
For some reason, the catch-all URLs did not work for me. I found that using the flask 404 handler results in the exact same thing. It sees the url and passes it down to react where your router will handle it.
#app.errorhandler(404)
def not_found(e):
return app.send_static_file('index.html')
Just to inform handle error 404 and render_template works perfectly for me.
#app.errorhandler(404)
def not_found(e):
return render_template("index.html")
I have to combine both catch-all and 404 handler for it to work properly. I am hosting a react-app in a subpath with its own redirection handler from react-router.
#app.route('/sub-path', defaults={'path': 'index.html'})
#app.route('/sub-path/<path:path>')
def index(path):
return send_from_directory('../react-dir/build', path)
#app.errorhandler(404)
def not_found(e):
return send_from_directory('../react-dir/build','index.html')
I am learning how to use Pluggable Views in Flask, since it seems that everyone uses them always for their advantages. I have the following file which returns an "Not Implemented Error". I am assuming that is because I am not implementing the dispatch_request. However, according to Flask's documentation, when using MethodView: "...if you implement a method called get() it means you will response to ’GET’ requests and the dispatch_request() implementation will automatically forward your request to that." Meaning, I do not require dispatch_request.
from flask import Flask, render_template, request, redirect, url_for, flash
from flask.views import View, MethodView
import os
SECRET_KEY = 'some_secret_key'
DEBUG = TRUE
app = Flask(__name__)
app.config.from_object(__name__)
class Main(View):
def dispatch_request(self):
return "Hello World!"
class Template(View):
def get(self):
return render_template('index.html')
def post(self):
result = eval(request.form['expression'])
flash(result)
return self.get()
app.add_url_rule('/', view_func=Main.as_view('main'))
app.add_url_rule('/template', view_func=Template.as_view('template'), methods=['GET', 'POST'])
if __name__ == "__main__":
app.run()
Oops.. silly Python beginner's mistake by me.
I was subclassing flask.views.View instead of flask.views.MethodView. flask.views.View requires dispatch_request, and does not automatically forward HTTP requests to dispatch_request as MethdoView does, hence the error.
I am trying to add a function in the Jinja environment from a blueprint (a function that I will use into a template).
Main.py
app = Flask(__name__)
app.register_blueprint(heysyni)
MyBluePrint.py
heysyni = Blueprint('heysyni', __name__)
#heysyni.route('/heysyni'):
return render_template('heysyni.html', heysini=res_heysini)
Now in MyBluePrint.py, I would like to add something like :
def role_function():
return 'admin'
app.jinja_env.globals.update(role_function=role_function)
I will then be able to use this function in my template. I cannot figure out how I can access the application since
app = current_app._get_current_object()
returns the error:
working outside of request context
How can I implement such a pattern ?
The message error was actually pretty clear :
working outside of request context
In my blueprint, I was trying to get my application outside the 'request' function :
heysyni = Blueprint('heysyni', __name__)
app = current_app._get_current_object()
print(app)
#heysyni.route('/heysyni/')
def aheysyni():
return 'hello'
I simply had to move the current_app statement into the function. Finally it works that way :
Main.py
from flask import Flask
from Ablueprint import heysyni
app = Flask(__name__)
app.register_blueprint(heysyni)
#app.route("/")
def hello():
return "Hello World!"
if __name__ == "__main__":
app.run(debug=True)
Ablueprint.py
from flask import Blueprint, current_app
heysyni = Blueprint('heysyni', __name__)
#heysyni.route('/heysyni/')
def aheysyni():
# Got my app here
app = current_app._get_current_object()
return 'hello'