We're using Flask for one of our API's and I was just wondering if anyone knew how to return a HTTP response 201?
For errors such as 404 we can call:
from flask import abort
abort(404)
But for 201 I get
LookupError: no exception for 201
Do I need to create my own exception like this in the docs?
You can use Response to return any http status code.
> from flask import Response
> return Response("{'a':'b'}", status=201, mimetype='application/json')
return {'a':'b'}, 201, {'Content-Type': 'application/json'}, or simply as return {'a': 'b'}, 201
{'a':'b'} gets automatically converted to {"a":"b"}. The double quotes are correct JSON. The original answer actually returns single quotes.You can read about it here.
return render_template('page.html'), 201
page.html... It won't work by itself.As lacks suggested send status code in return statement and if you are storing it in some variable like
notfound = 404
invalid = 403
ok = 200
and using
return xyz, notfound
than time make sure its type is int not str. as I faced this small issue also here is list of status code followed globally http://www.w3.org/Protocols/HTTP/HTRESP.html
Hope it helps.
status.NOT_FOUNDIn your flask code, you should ideally specify the MIME type as often as possible, as well:
return html_page_str, 200, {'ContentType':'text/html'}
return json.dumps({'success':True}), 200, {'ContentType':'application/json'}
...etc
Ripping off Luc's comment here, but to return a blank response, like a 201 the simplest option is to use the following return in your route.
return "", 201
So for example:
@app.route('/database', methods=["PUT"])
def database():
update_database(request)
return "", 201
you can also use flask_api for sending response
from flask_api import status
@app.route('/your-api/')
def empty_view(self):
content = {'your content here'}
return content, status.HTTP_201_CREATED
you can find reference here http://www.flaskapi.org/api-guide/status-codes/
In my case I had to combine the above in order to make it work
return Response(json.dumps({'Error': 'Error in payload'}),
status=422,
mimetype="application/json")
Dependent on how the API is created, normally with a 201 (created) you would return the resource which was created. For example if it was creating a user account you would do something like:
return {"data": {"username": "test","id":"fdsf345"}}, 201
Note the postfixed number is the status code returned.
Alternatively, you may want to send a message to the client such as:
return {"msg": "Created Successfully"}, 201
for error 404 you can
def post():
#either pass or get error
post = Model.query.get_or_404()
return jsonify(post.to_json())
for 201 success
def new_post():
post = Model.from_json(request.json)
return jsonify(post.to_json()), 201, \
{'Location': url_for('api.get_post', id=post.id, _external=True)}
You just need to add your status code after your returning data like this:
from flask import Flask
app = Flask(__name__)
@app.route('/')
def hello_world(): # put application's code here
return 'Hello World!',201
if __name__ == '__main__':
app.run()
It's a basic flask project. After starting it and you will find that when we request http://127.0.0.1:5000/ you will get a status 201 from web broswer console.
So, if you are using flask_restful Package for API's
returning 201 would becomes like
def bla(*args, **kwargs):
...
return data, 201
where data should be any hashable/ JsonSerialiable value, like dict, string.
Use something like this as per latest flask version 3.0.2
return render_template('main.html', status=201, data=data)#where main.html is template name and data is variables to be passed to the template
return '', 201