try:
mqtt_client.publish(feed, value, is_group=True)
except (ConnectionError) as e:
print('iotMqtt error in publish', e)
I get the following
TypeError: unexpected keyword argument 'is_group'
In this example, mqtt_client = adafruit_minimqtt.MQTT and feed="rhcudmore/feeds/nano1.state"
I tried with two versions of the circuit python libraries but get the same error?
adafruit-circuitpython-bundle-8.x-mpy-20231010
adafruit-circuitpython-bundle-8.x-mpy-20231025
I am running CircuitPython on a Arduino RP2040 connect.
Any suggestions on how to get is_group working?
p.s. in the source code for def publish()
if is_group:
self._client.publish("{0}/g/{1}".format(self._user, feed_key), data)
if shared_user is not None:
self._client.publish("{0}/f/{1}".format(shared_user, feed_key), data)
Even if is_group was working, what is this feed format with /g/? Rather than the more documented feed format with /f/.
I have a number of Arduinos, each publishes to their own group.
I see the publish()
is_group
optional parameter in the API documentation and in the source code.But when I call it from my CircuitPython code
I get the following
In this example,
mqtt_client = adafruit_minimqtt.MQTT
andfeed="rhcudmore/feeds/nano1.state"
I tried with two versions of the circuit python libraries but get the same error?
I am running CircuitPython on a Arduino RP2040 connect.
Any suggestions on how to get
is_group
working?p.s. in the source code for def publish()
Even if
is_group
was working, what is this feed format with/g/
? Rather than the more documented feed format with/f/
.