sed 无法仅打印匹配的正则表达式 grop

Question

I have some key value pair arguments. I need to print them as is. Example.

echo $X
(a=b) (c=d) (e=f)
echo "$X" | sed -E 's/([a-zA-Z0-9_]*=[a-zA-Z0-9_]*)/match/1'
echo "$X" | sed -E 's/([a-zA-Z0-9_]*=[a-zA-Z0-9_]*)/\1/1'
echo "$X" | sed -E 's/([a-zA-Z0-9_]*=[a-zA-Z0-9_]*)/\1/2'
echo "$X" | sed -E 's/([a-zA-Z0-9_]*=[a-zA-Z0-9_]*)/\1/3'

Post the above expresion, I wanted to print matching groups one by one. using.* in pattern matching is greedy and is printing either first or last matching groups only. How can I print any matching group in this way.

Here is my expected output.

a=b
c=d
e=f

Answer 1

This grep one-liner will do:

grep -o '[^(]*=[^)]*'

example:

kent$  grep -o '[^(]*=[^)]*' <<<'(a=b) (c=d) (e=f)' 
a=b
c=d
e=f

Answer 2

Replace ) ( with a newline and remove the remaining parentheses.

echo "$X" | sed 's/) (/\n/g;s/[()]//g'

To print the $n th line, you can pipe the output to

sed -n "$n p"