How can I print the last 4 columns using awk?
The following solution works fine with numbers:
Command:
echo "1 2 3 4 5" | awk '{print $(NF-3),$(NF-2),$(NF-1),$NF}'
Output:
2 3 4 5
However, when I use text with special characters like below, I get a syntax error:
echo "2023-01-03 01:05:06: Table test completed (0 hrs 0 mins 0.009 Secs)" | awk '{print $(NF-3),$(NF-2),$(NF-1),$NF}'
Output:
> " | awk '{print $(NF-2),$(NF-1),$NF}'
mins 0.009 Secs)
awk: The field -2 cannot be less than 0.
The input line number is 2.
The source line number is 1.
What I am expecting to print out is the below for the above example:
(0 hrs 0 mins 0.009 Secs)
How can I achieve this?
1st solution: With your shown samples please try following awk code. Simple explanation would be: Using match function of awk. Using regex \(.*\)$ which will match string from ( to till ) if match is found then printing sub-string which will print only matched value(s) as per OP's requirement.
echo "2023-01-03 01:05:06: Table test completed (0 hrs 0 mins 0.009 Secs" |
awk 'match($0,/\(.*\)$/){print substr($0,RSTART,RLENGTH)}' Input_file
2nd solution: If your Input_file always has "Table test completed" then you can make it more easy then 1st solution, try following then:
echo "Your_string" |
awk -F'[[:space:]]+Table test completed[[:space:]]+' '{print $NF}'