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Home/ Questions/Q 8591431
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T23:31:37+00:00 2026-06-11T23:31:37+00:00

I’d like to send a local REST request in a flask app, like this:

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I’d like to send a local REST request in a flask app, like this:

from flask import Flask, url_for, request
import requests

app = Flask(__name__)

@app.route("/<name>/hi", methods=["POST"])
def hi_person(name):
    form = {"name": name}
    return requests.post(url_for("hi", _external=True), data=form)

@app.route("/hi", methods=["POST"])
def hi():
    return 'Hi, %s!' % request.form["name"]

Sending curl -X POST http://localhost:5000/john/hi causes the entire flask app to freeze. When I send a kill signal, I get a broken pipe error. Is there a way to prevent flask from freezing here?

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  1. Editorial Team
    Editorial Team
    2026-06-11T23:31:38+00:00Added an answer on June 11, 2026 at 11:31 pm

    Run your flask app under a proper WSGI server capable of handling concurrent requests (perhaps gunicorn or uWSGI) and it’ll work. While developing, enable threads in the Flask-supplied server with:

    app.run(threaded=True)
    

    but note that the Flask server is not recommended for production use. As of Flask 1.0, threaded is enabled by default, and you’d want to use the flask command on the command line, really, to run your app.

    What happens is that using requests you are making a second request to your flask app, but since it is still busy processing the first, it won’t respond to this second request until it is done with that first request.

    Incidentally, under Python 3 the socketserver implementation handles the disconnect more gracefully and continues to serve rather than crash.

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