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

Off topic, this can be done using the grep also, just posting it here in case if anyone is looking for grep solution

echo 'xxx yyy zzze ' | grep -oE 'yyy'

Answer 2

If you know what column the text/pattern you're looking for (e.g. "yyy") is in, you can just check that specific column to see if it matches, and print it.

For example, given a file with the following contents, (called asdf.txt)

xxx yyy zzz

to only print the second column if it matches the pattern "yyy", you could do something like this:

awk '$2 ~ /yyy/ {print $2}' asdf.txt

Note that this will also match basically any line where the second column has a "yyy" in it, like these:

xxx yyyz zzz
xxx zyyyz