How to print matched regex pattern using awk?

Question

Using awk, I need to find a word in a file that matches a regex pattern.

I only want to print the word matched with the pattern.

So if

Answer 1

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.