Render a new template with socketio.on() in Flask

Question

I'm trying to do something along these lines: ``` from flask import Flask, render_template, redirect, url_for from flask.ext.socketio import SocketIO app = Flask(__name__) socketio = SocketIO(app) @app.route('/start') def start(): return render_template('start.html') @app.route('/new_view') def new_view(): return render_template('new_view.html') @socketio.on('change_view') def change_view(message): return redirect(url_for('new_view')) if __name__ == "__main__": socketio.run(app, host='127.0.0.1', port=8080) ``` Basically I want it to redirect if it gets the 'change_view' message from the client. Right now it gets to the `change_view()` function after I click a button that triggers the `socket.emit('change_view', message)` call, so that part works. It just doesn't redirect or get into the `new_view()` function at all (i.e. if I put a print statement in `new_view()` it doesn't print). But it also doesn't give me any errors. I am new to sockets so I'm guessing there's some fundamental mi…

Accepted Answer

Yeah, socket.io doesn't work like that. You can send a message telling the client to load a new page.

```
emit('redirect', {'url': url_for('new_view')})
```

Then in your client:

```
socket.on('redirect', function (data) {
    window.location = data.url;
});
```

But it's not clear why you need to hit the server at all for this particular example.

Render a new template with socketio.on() in Flask

Solution

Problem