Reputation: 1596
I'm using Swagger 2.0 to create the framework of a python-flask backend.
I declared the POST /image
endpoint as following on the Swagger online Editor:
/image:
post:
summary: "Description."
operationId: "image_post"
consumes:
- "multipart/form-data"
parameters:
- in: "formData"
name: "upfile"
description: "The file to upload."
required: true
type: "file"
responses:
201:
description: "Ok"
schema:
type: "string"
400:
description: "bad image"
x-swagger-router-controller: "swagger_server.controllers.default_controller"
After downloading the python-flask
server, I run it with the command python -m swagger_server
. When sending a post request through postman I get that the status code is 200
instead of 201
. It also shows 200
on the command line where the server is running.
Any clues if this is a bug? or am I doing something wrong?
Upvotes: 1
Views: 3165
Reputation: 555
I'm pretty sure you need to explicitly declare which status code you want to return in your view function; the swagger description is just documentation that your API server doesn't actually care about. Something like this pseudo code:
def image():
if request.method == 'POST':
if not is_valid(request.form):
return jsonify({'errors': [...]}), 400
# do stuff
return jsonify({'description': 'created'}), 201
Upvotes: 2