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Top Forums Shell Programming and Scripting grep based on pattern in a line and print the column before that Post 302659525 by balajesuri on Thursday 21st of June 2012 05:45:48 AM
Old 06-21-2012
Code:
awk -F'|' '{for(i=1;i<=NF;i++){ if($i == "reqUUID110612105016$BT") { print $(i-13) } }}' file.log

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GREP(1) 						      General Commands Manual							   GREP(1)

NAME
grep - search a file for lines containing a given pattern SYNOPSIS
grep [-elnsv] pattern [file] ... OPTIONS
-e -e pattern is the same as pattern -c Print a count of lines matched -i Ignore case -l Print file names, no lines -n Print line numbers -s Status only, no printed output -v Select lines that do not match EXAMPLES
grep mouse file # Find lines in file containing mouse grep [0-9] file # Print lines containing a digit DESCRIPTION
Grep searches one or more files (by default, stdin) and selects out all the lines that match the pattern. All the regular expressions accepted by ed and mined are allowed. In addition, + can be used instead of * to mean 1 or more occurrences, ? can be used to mean 0 or 1 occurrences, and | can be used between two regular expressions to mean either one of them. Parentheses can be used for grouping. If a match is found, exit status 0 is returned. If no match is found, exit status 1 is returned. If an error is detected, exit status 2 is returned. SEE ALSO
cgrep(1), fgrep(1), sed(1), awk(9). GREP(1)
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