How to print matched regex pattern using awk?

If Perl is an option, you can try this:

perl -lne 'print $1 if /(regex)/' file

To implement case-insensitive matching, add the i modifier

perl -lne 'print $1 if /(regex)/i' file

To print everything AFTER the match:

perl -lne 'if ($found){print} else{if (/regex(.*)/){print $1; $found++}}' textfile

To print the match and everything after the match:

perl -lne 'if ($found){print} else{if (/(regex.*)/){print $1; $found++}}' textfile

gawk can get the matching part of every line using this as action:

{ if (match($0,/your regexp/,m)) print m[0] }

match(string, regexp [, array]) If array is present, it is cleared, and then the zeroth element of array is set to the entire portion of string matched by regexp. If regexp contains parentheses, the integer-indexed elements of array are set to contain the portion of string matching the corresponding parenthesized subexpression. http://www.gnu.org/software/gawk/manual/gawk.html#String-Functions

This is the very basic

awk '/pattern/{ print $0 }' file

ask awk to search for pattern using //, then print out the line, which by default is called a record, denoted by $0. At least read up the documentation.

If you only want to get print out the matched word.

awk '{for(i=1;i<=NF;i++){ if($i=="yyy"){print $i} } }' file

It sounds like you are trying to emulate GNU's grep -o behaviour. This will do that providing you only want the first match on each line:

awk 'match($0, /regex/) {
    print substr($0, RSTART, RLENGTH)
}
' file

Here's an example, using GNU's awk implementation (gawk):

awk 'match($0, /a.t/) {
    print substr($0, RSTART, RLENGTH)
}
' /usr/share/dict/words | head
act
act
act
act
aft
ant
apt
art
art
art

Read about match, substr, RSTART and RLENGTH in the awk manual.

After that you may wish to extend this to deal with multiple matches on the same line.