Add error handling in Python API call for Meraki

I have this script to make an API call using the meraki python module to query a Meraki device.

I'm trying to add some error handling in the code so that if the API call comes back with an error code, it will do something else. I cannot seem to figure out what to do.

Here is my simple code to just query a device:

import meraki
import requests

API_KEY = 'API_KEY'
dashboard = meraki.DashboardAPI(API_KEY)

serial = input("What is the serial number?")
print(f"{serial}")

response = (dashboard.devices.getDevice(serial))

When I run the script it will return either a "200 OK" or "404 Not Found"

Terminal window response:

C:\Scripts\Meraki\dev> python .\getdevice.py
What is the serial number? XXXX-XXXX-XXXX

2023-04-26 18:32:52       meraki:     INFO > GET https://api.meraki.com/api/v1/devices/XXXX-XXXX-XXXX
2023-04-26 18:32:53       meraki:     INFO > devices, getDevice - 200 OK

2023-04-26 18:41:09       meraki:     INFO > GET https://api.meraki.com/api/v1/devices/XXXX-XXXX-XXXX
2023-04-26 18:41:10       meraki:    ERROR > devices, getDevice - 404 Not Found, b''

The purpose of this script is to check to see if the serial number has already been assigned to a user / network. If it is available, I'll get a "404 Not Found" and if it has already been assigned to someone, I'll get a "404 Not Found".

How can I detect and handle the 404 case?

Solution

I'm assuming you are using the meraki package with the source code at https://github.com/meraki/dashboard-api-python/tree/main/meraki.

According to that, if a 404 response occurs, getDevice should raise an APIError exception. It contains details about the error in the status, reason, and message attributes (see https://github.com/meraki/dashboard-api-python/blob/main/meraki/exceptions.py) – your log output shows the result of APIError.__repr__ which includes these attributes.

You can handle it by using a try/except statement:

from meraki import APIError

# ...

try:
    response = dashboard.devices.getDevice(serial)
except APIError as e:
    print(e.status)
    print(e.reason)
    print(e.message)

If you need to handle the 404 case specifically, you can use an if statement:

    if e.status == 404:
        # ...