Is there a way to replace characters from inside the regex?
like so:
find x | xargs perl -pi -e 's/(as dasd asd)/replace(" ","",$1)/'
From OP's comment
code find x | xargs perl -pi -e 's/work_search=1\/ttype=2\/tag=(.*?)">(.*?)<\/a>/work\/\L$1\E\" rel=\"follow\">$2<\/a>/g'
in this case i want $1's spaces be replaced with _
You can use a nested substitution:
$ echo 'foo bar baz' | perl -wpE's/(\w+ \w+ \w+)/ $1 =~ s# ##gr /e'
foobarbaz
Note that the /r modifier requires perl v5.14. For earlier versions, use:
$ echo 'foo bar baz' | perl -wpE's/(\w+ \w+ \w+)/my $x=$1; $x=~s# ##g; $x/e'
foobarbaz
Note also that you need to use a different delimiter for the inner substitution. I used #, as you can see.
As far as I understand, you want to remove the spaces. Is it correct?
You can do:
s/(as) (dasd) (asd)/$1$2$3/
code find x | xargs perl -pi -e 's/work_search=1\/ttype=2\/tag=(.*?)">(.*?)<\/a>/work\/\L$1\E\" rel=\"follow\">$2<\/a>/g' in this case i want $1's spaces be replaced with _