How do you extract IP addresses from a line matching a string from a file in bash?

Question

Example:

I have a file name example.txt and inside it this text:

some text here INFO    200     cv 58687 http://saomesitehoere.com live connect ASDFG 61.215.80.6 07:16

some text here INFO    100     fv 582702687 http://saomesitehoere.org live connect 31.15.80.1 07:16:33

some text here INFO    00     ov 587 http://saomesitehoere.uk live connect ASGGGGFG 91.211.80.6 09:16

some text here INFO    800    kcv 277 http://saomesitehoere.za live connect AFG 71.215.81.5 09:14

I want to extract the IP-address from the line which contain the string name "ASDFG", meaning 61.215.80.6

Anyone can help ?

Answer 1

$ grep -oP 'ASDFG \K\S*' < file
61.215.80.6

Answer 2

你可以尝试awk：

awk '/\<ASDFG\>/{print $(NF-1)}' file

Answer 3

您可以使用 ：

grep -oE '[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}'

Answer 4

An IP address is like three times one to three digits followed by a dot, followed by one to three digits. So this will give you the IP address in such a line:

$ grep ASDFG logfile | grep -o '\([[:digit:]]\{1,3\}\.\)\{3\}[[:digit:]]\+'
61.215.80.6

Answer 5

To extract all the "valid" IP addresses from the file:

gawk  '{match($0,/[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+/,a);split(a[0],b,".")} b[1]<=255&& b[2]<=255 && b[3]<=255 && b[4]<=255 &&length(a[0]){print a[0]}' input_file

This will work in two steps:

1. First it will extract all the sub-strings having pattern line digits.digits.digits.digits
2. it will check if each set of digits is less then or equal to 255.

Note: needs gawk to use match() function.