Get first word of match of last line - linux

Question

I want to parse through a log file formatted like this:
INFO: Successfully received REQUEST_ID: 1111 from 164.12.1.11
INFO: Successfully received REQUEST_ID: 2222 from 164.12.2.22
ERROR: Some error
INFO: Successfully received REQUEST_ID: 3333 from 164.12.3.33
INFO: Successfully received REQUEST_ID: 4444 from 164.12.4.44
WARNING: Some warning
INFO: Some other info
I want a script that outputs 4444. So extract the next word after ^.*REQUEST_ID: from the last line that contains the pattern ^.*REQUEST_ID.
What I have so far:
ID=$(sed -n -e 's/^.*REQUEST_ID: //p' $logfile | tail -n 1)
For lines match the pattern matches for, it deletes all the text matching the match thus leaving only the text after the match and prints it. Then I tail it to get the last line. How to do make it so it only prints the first word?
And is there a more efficient way of doing this then having it piped to tail?

Answer 1

With awk:
awk '
$4 ~ /REQUEST_ID:/{val=$5}
END {print val}
' file.csv
$4 ~ /REQUEST_ID:/ : Match lines in which Field # 4 match REQUEST_ID:.
{val=$5} : Store the value of field 5 in the variable val.
END {print val} : On closing the file, print the last value stored.
I have used a regex match to allow for some variance on the string, and yet get a match. A more lenient match will be (a match at any place of the line):
awk ' /REQUEST_ID/ {val=$5}
END {print val}
' file.csv
If you value (or need) more speed than robustness, then use (Quoting needed):
awk '
$4 == "REQUEST_ID:" {val=$5}
END {print val}
' file.csv

Answer 2

With GNU sed:
sed -nE 's/.* REQUEST_ID: ([0-9]+) .*/\1/p' | tail -n 1
Output:
4444
With GNU grep:
grep -Po 'REQUEST_ID: \K[0-9]+' file | tail -n 1
Output:
4444
-P: Interpret PATTERN as a Perl regular expression.
-o: Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
\K: Drop everything before that point from the internal record.

Answer 3

sed '/^.*REQUEST_ID: \([0-9]\{1,\}\) .*/ {s//\1/;h;}
$!d
x' ${logfile}
posix version
print an empty line if no occurence, the next word (assuming it's a number here)
Principe:
if line contain REQUEST_ID
extract the next number
put it in hold buffer
if not the end, delete the current content (and cycle to next line)
load holding buffer (and print the line ending the cycle)

Answer 4

You can match the number and replace with that value:
sed -e 's/^.*REQUEST_ID: \([0-9]*\).*$/\1/g' $logfile

Answer 5

Print field where line and column meet.
awk 'FNR == 5 {print $5}' file
4444

Answer 6

Another awk alternative if you don't know the position of the search word.
tac file | awk '{for(i=1;i<NF;i++) if($i=="REQUEST_ID:") {print $(i+1);exit}}'
yet, another one without looping
tac file | awk -vRS=" " 'n{print;exit} /REQUEST_ID:/{n=1}'