How to print only the first non-blank line using sed

Question

Given the following file, where line 1 is blank:

\n
Line 2\n
Line 3\n

How would you output only "Line 2" using sed?

Solutions using other standard UNIX tools, for example awk, are welcome.

Nowak

Answer 1

The examples work if you have an empty line, but not if you have a line containing characters like space or tab.

I think this version would work even if the "blank" line contains spaces or tabs.

sed '/[^[:blank:]]/q;d' file

Shelter

Answer 2

Using a version of grep with the -m switch, such as GNU or OpenBSD grep:

grep -m 1 . file

This stops reading the file after 1 matching line. . matches any character, so the first non-empty line will match.

Using any version of awk (essentially the same as the sed version):

awk '/./{print;exit}' file

Crony

awk 'NF{print;exit}' would also work depending on whether they count whitespace as blank or not. – Uncommercial

The non-standard -m extension was added to OpenBSD's grep(1) on Dec 9, 2017. Nice. Even more simple than sed. – Nowak

Answer 3

Multi-line version with comments

sed -n '  # use -n option to suppress line echoing
  /./ {   # match non-blank lines and perform the enclosed functions
          # print the non-blank line, i.e., "Line 2"
          p
          # quit right after printing the first non-blank line
          q
      }
' file

Single-line version without comments

sed -n '/./{p;q;}' file

Nowak

I would probably have said sed '/./q;d' file, but either works. – Cureall

@Wintermute, very nice. I did not realize the line would be echoed before quitting. – Nowak

It will if the -n option is not given to sed. q is similar to n in this regard. This behavior can be used nicely in this case, but to be honest, it is vexing more often than not. GNU sed has a Q command that quits without printing regardless of -n because of this. – Cureall