Shell Programming and Scripting

Hi all
I need to strip blank lines from the end of a file. I have searched and found topics on how to strip lines from the entirety of a file - however I need to limit this to only the last 3-4 lines.

Any ideas?

Thanks

Here's one solution using awk, from www.experts-exchange.com:

Code:

#!/usr/bin/awk -f

BEGIN { nb = 0; }

{
 if ( $0=="" )
 {
    nb++;
 }
 else
 {
   if( nb > 0 )
   {
     for(i = 0; i < nb; i++) print "";
     nb = 0;
   }
   print $0;
 }
}

Name the script something like yourScript and use it like this:

yourScript < yourFile > TMP_00
mv TMP_00 yourFile

Nice one, thanks.

Heres a little gizmo from my collection, its crude (no param checking) but it works




#!/bin/bash
#linechomp , removes trailing lines from end of file
numlines=`eval "cat $2 | wc -l"`
let splithere="$numlines-$1"
cat $2 | head -$splithere > $3
cat $2 | tail -$1 | sed -e '/^$/d' >> $3

usage: linechomp n foo bar

n is 3 or 4 in your case

Oomberas code might be better though as this doesn't guarantee you wont get

Line of some stuff, blah blah blah
<blank line>
Line of some more stuff
<blank line>

as your last 4 lines

Here's the same solution as oombera's though written in a more awk-like and less C-like fashion. Note that I'm also changing the problem somewhat: This will trim trailing lines consisting only of blanks and tabs (i.e., apparently empty) as well as truly empty lines:

Code:

#!/usr/bin/awk -f
NF == 0 { nb++ ; next }
nb      { for (i = 1; i <= NF; i++) print "" }
        { nb = 0; print }

Invoke it the same way as oombera's code. If you want only the truly empty lines to be whacked off the end, just change the first test to length == 0.