bash，仅从行首删除一半空格

Question

I need remove whitespaces from start of line, but only 1/2, for example:

    <div class="section" id="contact">
    <div class="container">
        <div class="col-md-12">
            <h4>04</h4>

to

  <div class="section" id="contact">
  <div class="container">
    <div class="col-md-12">
      <h4>04</h4>

etc.

Thanks in advance for any suggestion.

Answer 1

using group in sed

sed 's/^\([[:blank:]]\{1,\}\)\1/\1/' YourFile

you can use a white space instead of [[:blank:]] if you are sure it's space and not tab.

it do: replace a group of space by this group if it appear twice at the begining, so this is half the full group of space

Answer 2

You can try Perl

perl -pe  's!(^\s+)!$x=length($1)/2;" " x $x!sme ' input_file

with your given inputs

$ cat stanislav.txt
    <div class="section" id="contact">
    <div class="container">
        <div class="col-md-12">
            <h4>04</h4>

$ perl -pe  's!(^\s+)!$x=length($1)/2;" " x $x!sme ' stanislav.txt
  <div class="section" id="contact">
  <div class="container">
    <div class="col-md-12">
      <h4>04</h4>

$

Answer 3

Assuming all your leading white space is blank chars:

$ awk 'match($0,/^ */){$0=sprintf("%*s%s", int(RLENGTH/2), "", substr($0,RLENGTH+1))} 1' file
  <div class="section" id="contact">
  <div class="container">
    <div class="col-md-12">
      <h4>04</h4>

If not then run your file through pr -e -t first to make it so.