I need to search multiple patterns on the third column of a file, and then print the whole line.
I am using this one below but how do I get it to print the whole line where there is a match?
awk '{print $3}' file | egrep -w "S|M|D"
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please click edit and add few lines of input and expected output - some line should match and others shouldn't... this will help to add clarity to question as well as act as test data for those who wish to answer your questionSundeep– Sundeep2019-01-07 09:23:05 +00:00Commented Jan 7, 2019 at 9:23
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3 Answers
I think your requirement just needs awk and not a combination with grep. If you are looking to print the whole line where the third column matches any of those letters, you need to do
awk '$3 ~ /^(S|M|D)$/' file
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2since it is single character, you can also use
[SMD]Sundeep– Sundeep2019-01-07 09:28:37 +00:00Commented Jan 7, 2019 at 9:28 -
is this going to search for the exact match? like
grep -wdoes?hayabusa99– hayabusa992019-01-07 09:33:51 +00:00Commented Jan 7, 2019 at 9:33
To extract the lines whose 3rd whitespace-delimited field is exactly S, M or D, use one of
awk '$3 ~ /^[SMD]$/' file
or, using string matching rather than regular expression matching,
awk '$3 == "S" || $3 == "M" || $3 == "D"' file
A condition without a corresponding block will act as if its block simply was { print }.
awk '$3 ~ /^S/||/^M/||/^D/{print $0}' filename
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The way this parses out will match column 3 starting with an S, or column 1 starting with an M or D -- it misses column 3 M/D and it incorrectly includes the column 1 matches.2019-01-09 19:52:00 +00:00Commented Jan 9, 2019 at 19:52