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

Question

I have:

1 LINUX param1 value1 
2 LINUXparam2 value2
3 SOLARIS param3 value3
4 SOLARIS param4 value4

I need awk to print all lines in which $2 is LINUX.

Answer 1

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 2

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.

---

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

Answer 3

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

awk '$2~/LINUX/' file

Answer 4

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 5

I think it might be a good idea to include "exact" and "partial matching" cases using awk ))

So, for exact matching:

OTHER_SHELL_COMMAND | awk '$2 == "LINUX" { print $0 }'

And for partial matching:

OTHER_SHELL_COMMAND | awk '$2 ~ /LINUX/ { print $0 }'

Answer 6

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).