Python re.findall with multiple patterns

Question

I have a text file with entries like this:

 Interface01 :
     adress
        192.168.0.1
next-interface:
 interface02:
     adress
        10.123.123.214
next-interface:
 interface01 :
     adress
        172.123.456.123

I'd like to parse it and get only the IP address corresponding to Interface01

I tried may things with python re.finall but couldn't get anything matching

 i = open(f, r, encoding='UTF-8')
 txt = i.read()
 interface = re.findall(r'Interface01 :\s*(.adress*)n',txt,re.DOTALL)

but nothing works.

The expected result is 192.168.0.1.

Try Interface01\s*:\s*adress\s+(.*). Remove re.DOTALL. Use just re.search to get the first match. See ideone.com/QoE1uF. Can there be more IPs per interface? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Mar 22, 2018 at 13:48
Thanks for your help. @Maroun, The n is for the beginning of "new-interface". To Wiktor ok i try it now and tells you — ricardo
– ricardo, Commented Mar 22, 2018 at 13:52

Wiktor Stribiżew · Accepted Answer · 2018-03-22 13:59:13Z

3

You may use

Interface01\s*:\s*adress\s+(.*)

See the regex demo. In Python, use re.search to get the first match since you only want to extract 1 IP address.

Pattern details:

Interface01 - a literal substring
\s*:\s* - a : enclosed with 0+ whitespaces
adress - a literal substring
\s+ - 1+ whitespaces
(.*) - Group 1: any 0+ chars other than line break chars.

Python demo:

import re
reg = r"Interface01\s*:\s*adress\s+(.*)"

with open('filename') as f:
    m = re.search(reg, f.read())
    if m:
        print(m.group(1))

# => 192.168.0.1

answered Mar 22, 2018 at 13:59

Wiktor Stribiżew

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Comments

Hai Vu · Accepted Answer · 2018-03-22 14:00:03Z

2

How about creating a pattern that said "Interface01", then skip all chars that are not digits, then get the digits and dots?

re.findall(r'Interface01[^0-9]+([0-9.]+)', text)

Result:

['192.168.0.1']

Update

Thanks to @zipa, here is the updated regex:

re.findall(r'[iI]nterface01[^0-9]+([0-9.]+)', text)

Result:

['192.168.0.1', '172.123.456.123'

edited Mar 22, 2018 at 14:00

answered Mar 22, 2018 at 13:50

Hai Vu

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1 Comment

zipa Over a year ago

Small suggestion - change Interface into [Ii]nterface

Smart Manoj · Accepted Answer · 2018-03-22 13:50:01Z

0

interface = re.findall(r'Interface01 :\s*.adress\s*(.*?)$',txt,re.S|re.M)

answered Mar 22, 2018 at 13:50

Smart Manoj

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Comments

Austin · Accepted Answer · 2018-03-22 13:53:55Z

0

You could try something like this:

interface = re.findall(r'Interface01 :\n +adress\n +(\d+.\d+.\d+.\d+)', txt)
# ['192.168.0.1']

answered Mar 22, 2018 at 13:53

Austin

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Comments

RomanPerekhrest · Accepted Answer · 2018-03-22 13:55:13Z

0

For getting one single match it's better to use re.serach() function:

import re

with open('filename') as f:
    pat = r'Interface01 :\s*\S+\s*((?:[0-9]{1,3}\.){3}[0-9]{1,3})'
    result = re.search(pat, f.read()).group(1)

print(result)

The output:

192.168.0.1

answered Mar 22, 2018 at 13:55

RomanPerekhrest

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Comments

its me · Accepted Answer · 2018-03-22 14:00:11Z

0

you can use Interface01 :\n.*?\n(.*)

answered Mar 22, 2018 at 14:00

its me

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