Sign up ×
Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. It's 100% free, no registration required.

Replace text between brackets


testing on Linux [Remove white space] testing on Linux


testing on Linux [Removewhitespace] testing on Linux

So, how can we just remove all the white space between the brackets and achieve output as given ?

share|improve this question

marked as duplicate by Gilles, jasonwryan, uther, rahmu, Renan Feb 2 '13 at 13:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Remove comma between quotes only might help. – mtk Feb 1 '13 at 12:50
Replace text between brackets might also help. – manatwork Feb 1 '13 at 15:14

3 Answers 3

up vote 6 down vote accepted

If the [, ] are balanced and not nested, you could use GNU awk as in:

gawk -v RS='[][]' '
   NR % 2 == 0 {gsub(/\s/,"")}
   {printf "%s", $0 RT}'

That is use [ and ] as the record separators instead of the newline character and remove blanks on every other records only.

With sed, with the additional requirement that there be no newline character inside [...]:

sed -e :1 -e 's/\(\[[^]]*\)[[:space:]]/\1/g;t1'

If they are balanced but may be nested as in blah [blih [1] bluh] asd, then you could use perl's recursion regexp operators like:

perl -0777 -pe 's{(\[((?:(?>[^][]+)|(?1))*)\])}{$&=~s/\s//rsg}gse'

Another approach, which would scale to very large files would be to use the (?@{...}) perl regexp operator to keep track of the bracket depth like in:

perl -pe 'BEGIN{$/=\8192}s{((?:\[(?{$l++})|\](?{$l--})|[^][\s]+)*)(\s+)}

Actually, you can also process the input one character at a time like:

perl -pe 'BEGIN{$/=\1}if($l>0&&/\s/){$_=""}elsif($_ eq"["){$l++}elsif($_ eq"]"){$l--}'

That approach can be implemented with POSIX tools:

od -A n -vt u1 |
  tr -cs 0-9 '[\n*]' |
  awk 'BEGIN{b[32]=""; b[10]=""; b[12]=""} # add more for every blank
       !NF{next}; l>0 && $0 in b {next}
       $0 == "91" {l++}; $0 == "93" {l--}
       {printf "%c", $0}'

With sed (assuming no newline inside the [...]):

sed -e 's/_/_u/g;:1' -e 's/\(\[[^][]*\)\[\([^][]*\)\]/\1_o\2_c/g;t1' \
    -e :2 -e 's/\(\[[^]]*\)[[:space:]]/\1/g;t2' \
    -e 's/_c/]/g;s/_o/[/g;s/_u/_/g'

Are considered white space above any horizontal (SPC, TAB) or vertical (NL, CR, VT, FF...) spacing character in the ASCII charset. Depending on your locale, others might get included.

share|improve this answer
thank you i found a sed solution sed -e 's/([[^]]*)( )/\1/' – ekassis Feb 1 '13 at 15:35
@ekassis, not sure about your sed solution as it seems to have been mangled, but I don't expect it to remove more than one space inside [..] or handle more than one [...] per line. I've added a sed solution to my answer. – Stéphane Chazelas Feb 1 '13 at 15:44
you are right the sed -e 's/([[^]]*)( )/\1/' replace the last space between the brackets. the sed you provided me is the right one. THANK YOU – ekassis Feb 1 '13 at 17:20
I found this query when I was interested in remove spaces between double quotes. The first sed solution will only work when you're using a different char for opening and closing a pair (this issue may effect some of the other solutions provided also). – Sep 24 '14 at 14:30

Perl 5.14 solution (which is shorter and IMO easier to read—especially if you format it over multiple lines in a file, instead of as a one-liner)

perl -pE 's{(\[ .*? \])}{$1 =~ y/ //dr}gex'

That works because in 5.14, the regular expression engine is re-entrant. Here it is, expanded out and commented:

    (\[ .*? \])         # search for [ ... ] block, capture (as $1)
    $1 =~ y/ //dr       # delete spaces. you could add in other whitespace here, too
                        # d = delete; r = return result instead of modifying $1
}gex; # g = global (all [ ... ] blocks), e = replacement is perl code, x = allow extended regex
share|improve this answer
@StephaneChazelas well, doesn't assume anything about balanced. It does assume not nested, though. It works just fine on empty brackets (it leaves them alone, which is fine, as they contain no whitespace). It does assume one line (which would be easy to change). And indeed its only space, but trivial to change—as noted in the answer... – derobert Feb 1 '13 at 12:42
@StephaneChazelas Indeed, I suppose it does. Will fix. FYI, I don't mind if you edit my answers directly to fix things like that. – derobert Feb 1 '13 at 13:47

Perl solution:

perl -pe 's/(\[[^]]*?)\s([^][]*\])/$1$2/ while /\[[^]]*?\s[^][]*\]/'
share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.