How to print matched regex pattern using awk?

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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

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