Regular Expression to Capture specific set of IP Address in Python

Question

My output is as below

0.0.0.0/0
  unicast-ip4-chain
  [@0]: dpo-load-balance: [proto:ip4 index:86 buckets:1 uRPF:102 to:[0:0]]
    [0] [@0]: dpo-drop ip4
0.0.0.0/32
  unicast-ip4-chain
  [@0]: dpo-load-balance: [proto:ip4 index:87 buckets:1 uRPF:88 to:[0:0]]
    [0] [@0]: dpo-drop ip4
1.1.1.254/32
  unicast-ip4-chain
  [@0]: dpo-load-balance: [proto:ip4 index:72 buckets:1 uRPF:41 to:[0:0]]
    [0] [@5]: ipv4 via 2.2.2.254 VirtualFuncEthernet0/7/0.1540: mtu:1500 f8c001181ac0fa163e81a6c0810006040800
14.1.1.0/32
  unicast-ip4-chain
  [@0]: dpo-load-balance: [proto:ip4 index:14 buckets:1 uRPF:111 to:[0:0]]
    [0] [@0]: dpo-drop ip4
14.1.1.1/32
  unicast-ip4-chain
  [@0]: dpo-load-balance: [proto:ip4 index:54 buckets:1 uRPF:61 to:[1228:75011]]
    [0] [@5]: ipv4 via 14.1.1.1 VirtualFuncEthernet0/7/0.1540: mtu:1500 f8c001181ac0fa163e81a6c0810006040800

To capture the value 2.2.2.254 from the output I have written regexp as below.

var = 1.1.1.254/32
re.findall(var+r'.*ipv4\s+via\s+(\W+)', x1)

Current output is []

Unless you are specifically asking about how to solve a cross-version compatibility problem (in which case your question should obviously describe that problem) you should not mix the python-2.7 and python-3.x tags. I have edited to remove the former. — tripleee
– tripleee, Commented Jul 23, 2020 at 11:34

Wiktor Stribiżew · Accepted Answer · 2020-07-23 11:33:26Z

2

You may use

(?ms)^1\.1\.1\.254/32\n.*?ipv4\s+via\s+(\d[\d.]*)

See the regex demo

Details

(?ms) - re.M and re.DOTALL enabled
^ - start of a line
1\.1\.1\.254/32 - 1.1.1.254/32 string
\n - a newline
.*? - any 0 or more chars as few as possible
ipv4\s+via\s+ - ipv4, 1+ whitespaces, via, 1+ whitespaces
(\d[\d.]*) - Capturing group 1: a digit and then 0 or more digits / dots

Python demo:

import re
text = "0.0.0.0/0\n  unicast-ip4-chain\n  [@0]: dpo-load-balance: [proto:ip4 index:86 buckets:1 uRPF:102 to:[0:0]]\n    [0] [@0]: dpo-drop ip4\n0.0.0.0/32\n  unicast-ip4-chain\n  [@0]: dpo-load-balance: [proto:ip4 index:87 buckets:1 uRPF:88 to:[0:0]]\n    [0] [@0]: dpo-drop ip4\n1.1.1.254/32\n  unicast-ip4-chain\n  [@0]: dpo-load-balance: [proto:ip4 index:72 buckets:1 uRPF:41 to:[0:0]]\n    [0] [@5]: ipv4 via 2.2.2.254 VirtualFuncEthernet0/7/0.1540: mtu:1500 f8c001181ac0fa163e81a6c0810006040800\n14.1.1.0/32\n  unicast-ip4-chain\n  [@0]: dpo-load-balance: [proto:ip4 index:14 buckets:1 uRPF:111 to:[0:0]]\n    [0] [@0]: dpo-drop ip4\n14.1.1.1/32\n  unicast-ip4-chain\n  [@0]: dpo-load-balance: [proto:ip4 index:54 buckets:1 uRPF:61 to:[1228:75011]]\n    [0] [@5]: ipv4 via 14.1.1.1 VirtualFuncEthernet0/7/0.1540: mtu:1500 f8c001181ac0fa163e81a6c0810006040800"
v = "1.1.1.254/32"
m = re.search(rf"^{re.escape(v)}\n.*?ipv4\s+via\s+(\d[\d.]*)", text, re.M|re.DOTALL)
if m:
    print(m.group(1))
# => 2.2.2.254

answered Jul 23, 2020 at 11:33

Wiktor Stribiżew

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2 Comments

Raj Naik Over a year ago

how can we capture the multiple IP, if I have to capture multiple IP. eg :

1.1.1.254/32   unicast-ip4-chain   [@0]: dpo-load-balance: [proto:ip4 index:214 buckets:4 uRPF:230 to:[0:0]]     [0] [@5]: ipv4 via 2.2.2.254 Interface0/6/0.2000: mtu:1500 f8c001189ac0fa163e77d390810005e40800     [1] [@5]: ipv4 via 3.3.3.254 Interface0/6/0.2001: mtu:1500 f8c001189ac0fa163e77d390810005e50800     [2] [@5]: ipv4 via 4.4.4.254 Interface0/6/0.2002: mtu:1500 f8c001189ac0fa163e77d390810006300800     [3] [@5]: ipv4 via 5.5.5.254 Interface0/6/0.2003: mtu:1500 f8c001189ac0fa163e77d390810006310800

Raj Naik Over a year ago

Written a regexp as below as below for the same, to capture 2.2.2.54, 3.3.3.254, 4.4.4.254, 5.5.5.254 Python Code for the same

Jacek Rojek · Accepted Answer · 2020-07-23 11:47:36Z

0

If you want the whole block of text until next IP try this: https://regex101.com/r/wCHYvj/2

(1\.1\.1\.254\/32)(\s*|.*)+?(?=\d{1,3}.\d{1,3}.\d{1,3})

find given IP and capture until next IP in new line

answered Jul 23, 2020 at 11:47

Jacek Rojek

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2 Comments

Wiktor Stribiżew Over a year ago

(\s*|.*)+? will lead to huge performance issues and catastrophic backtracking with non-matching inputs.

Jacek Rojek Over a year ago

@WiktorStribiżew good point, thx. I'll edit my answer

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