Print or grep lines which start from second column

Question

I want to print lines which start from second column (first column is white space).

$ cat test.txt

first
 second
  third
   four
 second
    five
      seven

Output should be:

 second


 second

When the line starts from the 7th column then the output should be:

      seven

Answer 1

You can use ^\s\S regex:

$ cat test.txt
first
 second
  third
   four
 second
    five
      seven
$ grep "^\s\S" test.txt
 second
 second

• ^ stands for "line starts with",
• \s means "first character is a white space",
• \S for "following character is not a whitespace.

This way, you get every line beginning with a single whitespace.

If you want lines beginning with n whitespaces, add {n}:

$ grep "^\s\{1\}\S" test.txt
 second
 secon d
$ grep "^\s\{2\}\S" test.txt
  third
$ grep "^\s\{3\}\S" test.txt
   four
$ grep "^\s\{4\}\S" test.txt
    five
$ grep "^\s\{5\}\S" test.txt
$ grep "^\s\{6\}\S" test.txt
      seven

Answer 2

Really simple solution for the second version:

 echo "first
 second
  third
   four
 second
    five
      seven" | grep -e '^ ' | grep -ve '^  '

[Edit] - now working with only the withespaces at the begining of the line