[SOLVED] Question with Netmiko and TextFSM against cisco IOSXE

Question with Netmiko and TextFSM against cisco IOSXE

Trying to do a check against Cisco IOSXE to see if any ports are still configured in the default VLAN.

Output = net_connect.send_command ('show int status', use_textfsm=true)
for i in output:
  if i["vlan"] == "1":
    print ('Not compliant')
  else:
    print ('Compliant')

This does work, but for a 48 port switch, I get 48 lines saying Compliant or Not Compliant. How can I change this so that if all ports are in a different vlan, lets say vlan 2, I get one line saying Complaint. And if ANY number of ports are in VLAN 1 , whether is be 1 port or 10 ports, I get one line saying "Not complaint", instead of 48 lines.

Solution

You have many options here. I will try to name some:

Move for loop into a function

def check_complaint(output):
  for i in output:
    if i["vlan"] == "1":
      return False
  return True

output = net_connect.send_command ('show int status', use_textfsm=true)
if check_complaint(output):
  print("Compliant")
else:
  print("Not Compliant")

Using the return statement in the functions allows the code to stop at the first interface in vlan 1

Use a variable

output = net_connect.send_command ('show int status', use_textfsm=true)
compliant = True
for i in output:
  if i["vlan"] == "1":
    compliant = False
    break

if compliant:
    print("Compliant")
else:
    print("Not Compliant")

Thanks to the break statment we will exit the for loop at the first vlan 1 match.

Using any with a generator expression

output = net_connect.send_command ('show int status', use_textfsm=true)
if any(i["vlan"] == "1" for i in output):
    print("Not Compliant")
else:
    print("Compliant")

The any function will also stop at the first match of Vlan 1.

The code is not tested and not perfect but should give you some ideas how to solve your problem