How to extract the characters I need from a multiline string-CodePudding

I need to display the ip address after address: . Everything else needs to be trimmed. What code will most optimally solve my problem?

show interface PPPoE0

               id: PPPoE0
            index: 0
             type: PPPoE
      description: Internet (NetFriend)
   interface-name: PPPoE0
             link: up
        connected: yes
            state: up
              mtu: 1400
         tx-queue: 1000
          address: 46.42.50.121
             mask: 255.255.255.255
           global: yes
        defaultgw: yes
         priority: 1000
   security-level: public
        auth-type: PAP, CHAP, MS-CHAP, MS-CHAPv2
           remote: 46.42.48.1
           uptime: 45562
       session-id: 23430
             fail: no
              via: GigabitEthernet0/Vlan2
      last-change: 45562.183918

(config)> exit
Core::Configurator: Bye.

CodePudding user response：

You can use a simple Regular Expression to search for all matching IPs in a string.

The regex query for an IP Adress would be : (\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})
If you want to learn what the RegEx query does you can see the full explanation here.

The code would look something like

import re

text = """
        show interface PPPoE0
    
                   id: PPPoE0
                index: 0
                 type: PPPoE
          description: Internet (NetFriend)
       interface-name: PPPoE0
                 link: up
            connected: yes
                state: up
                  mtu: 1400
             tx-queue: 1000
              address: 46.42.50.121
                 mask: 255.255.255.255
..................
"""

regex_pattern_ip = re.compile(r'(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})')

ip = regex_pattern_ip.search(text)[0]

print(ip)

Hope this helped

CodePudding user response：

I don't know about most optimal, but one way to do it is to split by lines, find the one containing address, and extract the text from it:

def getIp(text):
    lines = text.split("\n")
    for line in lines:
       line = line.replace(" ").replace("\t")#in case it has newlines or spaces in front, not sure based off question
       if line.startswith("address:"):
           return line[8:]
    raise Exception("Address line not found")

CodePudding user response：

Maybe not optimal, but will work:

print(s[s.find("address: ")   9: s.find("mask")].strip())