grep or sed for word containing string

Question

example file:

blahblah 123.a.site.com   some-junk
yoyoyoyo 456.a.site.com   more-junk
hihohiho 123.a.site.org   junk-in-the-trunk
lalalala 456.a.site.org   monkey-junk

I want to grep out all those domains in the middle of each line, they all have a common part a.site with which I can grep for, but I can't work out how to do it without returning the whole line?

Maybe sed or a regex is need here as a simple grep isn't enough?

Answer 1

You can do:

grep -o '[^ ]*a\.site[^ ]*' input

or

awk '{print $2}' input

or

sed -e 's/.*\([^ ]*a\.site[^ ]*\).*/\1/g' input

Answer 2

Try this to find anything in that position

  $ sed -r "s/.* ([0-9]*)\.(.*)\.(.*)/\2/g"

 [0-9]* - For match number zero or more time.
 .*     - Match anything zero or more time.
 \.     - Match the exact dot.
 ()     - Which contain the value particular expression in parenthesis, it can be printed using \1,\2..\9. It contain only 1 to 9 buffer space. \0 means it contain all the expressed pattern in the expression.