Flask and React routing

Question

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.

Answer 1

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.

Answer 2

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')