I have created a simple flask app that and I'm reading the response from python as:
response = requests.post(url,data=json.dumps(data), headers=headers )
data = json.loads(response.text)
Now my issue is that under certain conditions I want to return a 400 or 500 message response. So far I'm doing it like this:
abort(400, 'Record not found')
#or
abort(500, 'Some error...')
This does print the messa on the terminal:
But in the API response I keept getting a 500 error response:
EDIT:
The structure of the code is as follows:
|--my_app
|--server.py
|--main.py
|--swagger.yml
Where server.py has this code:
from flask import render_template
import connexion
# Create the application instance
app = connexion.App(__name__, specification_dir="./")
# read the swagger.yml file to configure the endpoints
app.add_api("swagger.yml")
# Create a URL route in our application for "/"
@app.route("/")
def home():
"""
This function just responds to the browser URL
localhost:5000/
:return: the rendered template "home.html"
"""
return render_template("home.html")
if __name__ == "__main__":
app.run(host="0.0.0.0", port="33")
And main.py has all the function I'm using for the API endpoints.
E.G:
def my_funct():
abort(400, 'Record not found')
When my_funct is called, I get the 'Record not found' printed on the terminal, but not in the response from the API itself, where I always get the 500 message error.
You have a variety of options:
The most basic:
@app.route('/')
def index():
return "Record not found", 400
If you want to access the headers, you can grab the response object:
@app.route('/')
def index():
resp = make_response("Record not found", 400)
resp.headers['X-Something'] = 'A value'
return resp
Or you can make it more explicit, and not just return a number, but return a status code object
from flask_api import status
@app.route('/')
def index():
return "Record not found", status.HTTP_400_BAD_REQUEST
You can read more about the first two here: About Responses (Flask quickstart)
And the third here: Status codes (Flask API Guide)