How to use awk to print lines where a field matches a specific string?

In awk:

awk '$2 == "LINUX" { print $0 }' test.txt

See awk by Example for a good intro to awk.

In sed:

sed -n -e '/^[0-9][0-9]* LINUX/p' test.txt

See sed by Example for a good intro to sed.

My answer is very late, but no one has mentioned:

awk '$2~/LINUX/' file

This is a case in which you can use the beautiful idiomatic awk:

awk '$2=="LINUX"' file

That is:

• The default action of awk when in a True condition is to print the current line.
• Since $2 == "LINUX" is true whenever the 2nd field is LINUX, this will print those lines in which this happens.

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In case you want to print all those lines matching LINUX no matter if it is upper or lowercase, use toupper() to capitalize them all:

awk 'toupper($2)=="LINUX"' file

Or IGNORECASE with either of these syntaxs:

awk 'BEGIN {IGNORECASE=1} $2=="LINUX"' file
awk -v IGNORECASE=1 '$2=="LINUX"' file