awk：逐行读取文件并sed以解析列中的IP

Question

I have an issue using awk and sed to parse a log file.

The below file:

DDDD;callID:4565;AAAA;Body:test message 1;BBBB;ng?10.1.5.60 4565;complete
DDDD;callID:5489;AAAA;Body:test message 2;BBBB;ng?10.4.100.3 5489;complete
DDDD;callID:3456;AAAA;Body:test message 3;BBBB;ng?10.200.1.5 3456;failed

Using the following command works fine to strip out IP from column 7. I am relying on removing last 5 characters from end to output only IP.

`awk -F";" ' {print $2 ";"  $4 ";" $7  ";" $6}' test.CSV | sed -e 's/Body://g; s/callID://g; s/ng?//g' | sed 's/.\{5\}$//'  >> complete.txt`

output as expeted:

4565;test message 1;complete;10.1.5.60
5489;test message 2;complete;10.4.100.3
3456;test message 3;failed;10.200.1.5

9 out of 10 times this is fine, but just found out column 7 will not always contain 4 digits at the end. Column 7 contains user input data so if something larger than 4 digit is entered incorrectly, it does not strip the IP as needed. How can I remove everything but the IP for that column?

Example of incorrect field:

DDDD;callID:3456;AAAA;Body:test message 3;BBBB;ng?10.200.1.5 3456345;failed

bad output:

3456;test message 3;failed;10.200.1.5 34

I was trying to use

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

inside the {print.....} section but could not get it to work.....

thank you in advance for your help.

Answer 1

$ awk -F'[:;?]' -v OFS=';' '{sub(/ .*/,"",$9); print $3,$6,$10,$9}' file
4565;test message 1;complete;10.1.5.60
5489;test message 2;complete;10.4.100.3
3456;test message 3;failed;10.200.1.5

Answer 2

Using awk

awk '{split($2,a,":");split($4,b,":");split($6,c,"[? ]");print a[2],b[2],$7,c[2]}' FS=";" OFS=";" file

4565;test message 1;complete;10.1.5.60
5489;test message 2;complete;10.4.100.3
3456;test message 3;failed;10.200.1.5