Question

Just now I was looking for a regex that will detect string not enclosed in double quotes. I got fabulous answer to that which worked in javascript. I am actually trying to run it using grep but it isn’t working. Can anyone help me why this regex isn’t working

Sample string : "quick" "brown" fox jumps "over" "the" lazy dog

I am trying to run : echo "quick" "brown" fox jumps "over" "the" lazy dog | grep -e '(?<![\S"])([^"\s]+)(?![\S"])'

It gives no output. How different are linux regexes from normal regexes. How can this be done?
Please help. Refer to this link: http://regexr.com?31eg3 for the same

Answer 1

Two things, first you need to quote the whole string to echo, or the shell strips the double quotes. Then you are using perl-style regular expressions, so you need to tell grep to use them for matching with the -P option.

$ echo "quick" "brown" fox jumps "over" "the" lazy dog
quick brown fox jumps over the lazy dog
$ echo "quick" "brown" fox jumps "over" "the" lazy dog | grep -e '(?<![\S"])([^"\s]+)(?![\S"])'
$ echo '"quick" "brown" fox jumps "over" "the" lazy dog' | grep -e '(?<![\S"])([^"\s]+)(?![\S"])'
$ echo '"quick" "brown" fox jumps "over" "the" lazy dog' | grep -P -o -e '(?<![\S"])([^"\s]+)(?![\S"])'
fox
jumps
lazy
dog

Here I added -o, so it is clear what was matched.