awk and/or sed: Only print lines where the second field matches some criteria

Question

I have

1 LINUX param1 value1

2 LINUXparam2 value2

3 SOLARIS param3 value3

4 SOLARIS param4 value4

need by awk to pring all lines that $2 is LINUX THX

Answer 1

Try these out:

egrep -i '^\w+ LINUX ' myfile

awk '{IGNORECASE=1}{if ($2 == "LINUX") print}' myfile

sed -ne '/^[0-9]* [Ll][Ii][Nn][Uu][Xx] /p' myfile

edit: modified for case insensitivity

Answer 2

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.

Answer 3

In GNU sed case-insensitive matches can be made using the I modifier:

sed -n '/^[^[:space:]][[:space:]]\+linux[[:space:]]\+/Ip'

Will robustly match "linux", "Linux", "LINUX", "LiNuX" and others as the second field (after the first field which may be any non-whitespace character) and surrounded by any amount (at least one) of any whitespace (primarily space and tab, although you can use [:blank:] to limit it to strictly those).