parsing file in bash

Question

I have a output in a file and i want to parse the file line by line to just get the line with the "ip address(10.113.193.70)" and the "Version: 47.80.99.08" to the standard output. What would be the best way to parse using grep/awk or bash?

10.113.193.70

Executing (0,0,0,0,0,0,0,0,0,0):
==== product information ====
Product: e10
Package: default
Version: 47.80.99.08
Type:    customer
Builder: symsm
Project: RAID

Answer 1

all of awk sed grep bash will work

this is assuming that the line with the IP is on the start of the line, and no other line starts with a digit.
Otherwise you can refine the IP pattern.

bash:

#!/usr/bin/env bash
while read -r
do [[ $REPLY =~ ^(Version|[[:digit:]]) ]] && echo "$REPLY"
done < file

grep:

$ grep "^\(Version\|[[:digit:]]\)" file
10.113.193.70
Version: 47.80.99.08

awk:

$ awk '$1 ~ /^(Version|[0-9])/ { print }' file
10.113.193.70
Version: 47.80.99.08

sed:

$ sed -n '/^\(Version\|[[:digit:]]\)/p' file
10.113.193.70
Version: 47.80.99.08

Answer 2

One approach:

grep '[0-9][.][0-9]*[.][0-9]*[.][0-9]' filename