grep - remove lines which only contain whitespace.
 

I'm having some trouble getting grep to remove lines from a file which only contain white space characters.

The attached file contains lines which are commented out by having a # at the start, lines which are empty and lines which contain either just some spaces or tab characters. I want to use grep to remove all the aforementioned types of line and end up with lines 2 3 6 13.

I thought that this would work:

grep -ve ^# lines.txt | grep -ve ^$ | grep -ve '^\s*$'

But it doesn't. The first and second greps do what I expect. The first one removes the commented out lines, the second removes the empty lines. The last grep, which I think should remove lines containing only whitespace characters doesn't remove anything and I can't figure out what the correct regexp should be. Anyone know?

Hi,

egrep -v "^[ ]*$|^#" lines.txt

There is a space and a tab between the square brackets. A tab can be 'cerated' by: ctrl-v then tab

Hope this helps.

Whilst I appreciate the suggestion, I really don't like that as a solution because of the invisibility of the tab character. It would be far too easy to forget that it's there, or to lose it in a copy/paste. I tried using

But that doesn't work.

Hi again,

I don't get the objection:

1) If you use this as a oneliner you type as you go and the oneliner is forgotten after you use it (getting it back from your history is a possibility).

2) If this is used in a script: A comment above the command should solve the forgetting part.

Anyway, here's another solution (but more typing... ;) ):

egrep -v "^[[:space:]]*$|^#" lines.txt

That's great, thanks a lot.