I have a link and I would like to return only content between www. and .com
e.g www.blablabla.com would return only blablabla
How could I do that? When I use grep '\.[a-zA-Z0-9\.-]*\.' it gives me .blablabla.
2 Answers
$ echo "www.blablabla.com" | grep -oP '(?<=\.)[a-zA-Z0-9\.-]*(?=\.)' blablabla -o -- print only matched parts of matching line
-P -- Use Perl regex
(?<=\.) -- after a literal ., aka, a "positive look-behind" ...
[a-zA-Z0-9\.-]* -- match zero or more instances of lower & upper case characters, numbers 0-9, literal . and hyphen ...
(?=\.) -- followed by a literal ., aka a "positive look-ahead"
See this link for more on look arounds. Tools like https://regex101.com/ can help you break down your regular expressions.
- Thanks that's what i wanted but what does it do ? Could u explain it a bit more please? Also
-Puses Perl regular expression is there any way to do it just with grep regular expressions?pnom– pnom2016-05-02 18:13:50 +00:00Commented May 2, 2016 at 18:13 -
- 2If you don't want to use
-Pthere is no way you can do this using grep re alone. If you want to stick withgrepconsider usingtrto drop the.like thisecho 'www.blablabla.com' | grep -o '\.[a-zA-Z0-9\.-]*\.' | tr -d .GMaster– GMaster2016-05-03 01:34:42 +00:00Commented May 3, 2016 at 1:34
sed solution:
$ str='Hellowww.hello.comMywww.world.comWorld' $ echo "$str" | sed -e 's/com/com\n/g' | sed -ne '/.*www\.\(.*\)\.com.*/{ s//\1/p }' hello world 
awk -F. '{print $2}'.." Apparently, you don't wantgrep -Peither. To get the best answer, you should explain your requirements. There are many good solutions to this problem. You should explain why you are rejecting the best ones.