How to display the header of a matched row in a file?

Shell Programming and Scripting
1

Hi,

So I am trying to print the first row(header) first column alongwith the matched value. But I am not sure how do I print the same, by matching a pattern located in the file

eg
File contents

Name Place 
Jim     NY
Jill      NJ
Cathy CA
Sam   TX
Daniel FL

And what I want is to print the first column first row(i.e header) for Place matching CA

and the matched value row

o/p is

Name
Cathy CA

So I know with awk

 awk /CA/ File

will give me just

Cathy CA

How do I get the first column title "Name"?

2

Perhaps something like:

/usr/xpg4/bin/awk -v place="CA" '
FNR == 1 {
	print $1
	next
}
$2 == place' File
3

Sidnow,

Are you looking for:

awk 'NR == 1 { print $1; } /CA/ { print; }'