[Solved] What is wrong with this Flask health check endpoint code? from flask import Flask, jsonify app = Flask(__name__) @app.route('/health') def health(): return jsonify(status='ok'), 201 | Flask

Flask - Deployment

What is wrong with this Flask health check endpoint code?

from flask import Flask, jsonify
app = Flask(__name__)

@app.route('/health')
def health():
    return jsonify(status='ok'), 201

A201 status code is inappropriate for health check

Bjsonify cannot be used with status codes

CRoute decorator syntax is invalid

DFunction must return a string, not jsonify

Step-by-Step Solution

Solution:

Step 1: Understand HTTP status codes for health checks
Health checks typically return 200 OK, not 201 Created which means resource creation.
Step 2: Confirm jsonify usage and syntax
jsonify can be returned with a status code tuple; syntax is valid.
Final Answer:
201 status code is inappropriate for health check -> Option A
Quick Check:
Use 200 OK for health checks, not 201 [OK]

Quick Trick: Use 200 OK status for health checks [OK]

Common Mistakes:

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What is wrong with this Flask health check endpoint code?