Replace possible blank space at the beginning of the line

Question

So I try to replace blank space at the beginning of the line like so.

Original :

this
 is

\tsome
  text

What I want :

+this
+is
+
+some
+ text

I tried to use this regex to do that :

echo -e 'this\n is\n\n\tsome\n  text' |
perl -wpe 's/^\s?(.*)/+$1/'

But this is the result for that :

+this
+is
++some
+ text

So my line with no text at all is not processed correctly. Or rather, I don't understand why it behaves like that.

Any idea why ?

Thanks !

Maybe you can use `\h` instead of `\s` (see https://stackoverflow.com/questions/3469080/match-whitespace-but-not-newlines). — sticky bit, Apr 07 '19 at 15:57

ikegami · Answer 1 · 2019-04-08T18:04:35.570

You are replacing the line-ending Line Feed. One solution is to remove it before the match and re-add it afterwards. This can be done using -l.

perl -ple's/^\s?/+/'

Alternatively, you could remove the Line Feed from the set of characters you are matching.

perl -pe's/^[^\S\n]?/+/'

If you're ok with also leaving VT, FF, CR, LINE SEPARATOR and PARAGRAPH SEPARATOR in place, you can shorten the above to the following:

perl -pe's/^\h?/+/'

score 2 · Answer 2 · answered Apr 07 '19 at 16:29

2

As pointed out by sticky bit in the comment, I just have to replace \s by \h. This is solved !

answered Apr 07 '19 at 16:29

ogr

1

Also, you don't need to capture the rest of the line: `s/^\h?/+/` is enough – Dada Apr 07 '19 at 17:17

2 Answers2