UNIX for Beginners Questions & Answers

sed -n                         # -n: Do not automatically print records
/CartService/                  # Look for lines that contain "CartService"
s/.*\(..:..:........\).*/\1/p  # Discard everything but the last pattern on the line that looks like:
                               # 2 chars - colon - 2 chars - colon - 6 chars 
                               # (the .* gobbles up the first pattern on the line)
                               # \1 is a so-called back reference, it refers to the pattern within the escaped brackets
                               # So what it all does - in short - is look for "CartService" and then
                               # replace the entire line with the pattern and if it can do that, print the line

awk 'match($0,/UA:MP:[a-zA-Z]+[0-9]+/) && !a[substr($0,RSTART,RLENGTH)]++{sub(/\[/,"",$1);sub(/\]/,"",$2);print $1,$2,substr($0,RSTART,RLENGTH-2)}'   Input_file

awk '
match($0,/UA:MP:[a-zA-Z]+[0-9]+/) && !a[substr($0,RSTART,RLENGTH)]++{
   sub(/\[/,"",$1);
   sub(/\]/,"",$2);
   print $1,$2,substr($0,RSTART,RLENGTH-2)
}'   Input_file

zgrep -B1 "Failed to calculate US Tax" log-gr_base.log.2018-05-08.gz | grep "CartService" | grep -Eo '.{4}-.{2}-[^]]*|[[:upper:]]{2}:[[:upper:]]{2}:[[:upper:]]{2}[[:digit:]]{6}'

zgrep -B1 "Failed to calculate US Tax" log-gr_base.log.2018-05-08.gz | grep "CartService" | grep -Eo '.{4}-.{2}-[^]]*|.{2}:.{2}:.{8}'

zgrep -B1 "Failed to calculate US Tax" log-gr_base.log.2018-05-08.gz | grep "CartService" | grep -Eo '.{4}-.{2}-[^]]*|.{2}:.{2}:.{8}' | paste - -

sort -u -k3,3

gunzip -c log-gr_base.log.2018-05-08.gz | awk '/Failed to calculate US Tax/ && !A[p]++{print q OFS p} {split ($0,F,/[][]/); p=$5; q=F[2]}'