Sed filter words from string

Shell Programming and Scripting
1

I have following string in a variable:

str="sstring garbage adfsdf tab.col1 lkad rjfj tab2.col2 adja tab2.col4 garbage"

I want to filter "word.word" pattern from it. I assume the pattern won't occur in start or end of the string. Output shoud be:

tab.col1 tab2.col2 tab2.col4

Can this be done through sed? My UNIX flavor does not have any GNU extensions:

Thanks in advance.

2 
echo "sstring garbage adfsdf tab.col1 lkad rjfj tab2.col2 adja tab2.col4 garbage" |grep -o "[[:graph:]]*\.[[:graph:]]*"
3

I don't know about sed, but here's one way to do it with Perl -

$ 
$ 
$ 
$ echo "sstring garbage adfsdf tab.col1 lkad rjfj tab2.col2 adja tab2.col4 garbage" | perl -ne '
  while(/[^.]* ([^. ]+)\.([^. ]+)/g){print "$1.$2 "} END {print "\n"}'
tab.col1 tab2.col2 tab2.col4 
$ 
$

tyler_durden

4 
for i in $str; do case $i in *?.?*) echo $i; esac; done

---------- Post updated at 10:27 ---------- Previous update was at 10:17 ----------

I think sed becomes too complicated for this purpose, e.g.:

pat="[^ \t]\{1,\}\.[^ \t]\{1,\}"
echo $str | sed "s|$pat|&\n|g" | sed -n "s|.*[ \t]\($pat\).*|\1|p"
5

Another one with awk:

echo $str | awk '/\./' RS=" "
6

Nice and short, but that would also filter out the dot at the end of a sentence, no?

7

IMHO the OP wants words with a dot between 2 words, no?

8

Yes.. Well I was able to do through sed by replacing the spaces in the string with \n and then applying the search pattern. Thanks for the help.