Chopping off the last field of each line?

well done

try this

awk '{$NF=""; print $0}'

Cheers
Amit

I prefer:

awk '{NF--;print}'

With NF="" you will have additional spaces at the end of each line.

Nice, this just solved a problem I've been working on. So thanks.

I figured I'd try to make it work without the rev. Reversing the argument of the -f parameter should returned the same results.

awk '{NF--;print}' Prints the whole name for me. What am I missing?

awk '{NF="";print}' works, but I need to include the periods.

I had to strip the last field from some 2Million line apache logs. First I tried the reverse-cut-reverse approach and that was likely to take 'forever'.
So I found a solution with sed that worked a lot better:
date; time (gzcat infile.gz | sed 's/\(.*\)\ \(.*\)/\1/' | gzip -c - > outfile.gz ); date
What the sed command does is strip 1 parameter to the preceeding space from the end of the line.

I managed to do this with awk:

cat filename |awk -F/ '{gsub($NF,"");print}'

The gsub replaces the last field with a null string. My delimiter is /

Nick

Thanks Nick..

awk -F/ '{gsub($NF,"");print}'

helped me too much.. was searching for this..

I am adding something else

awk -F/ '{gsub($NF,"");sub(".$", "");print}'

to cut the last part with delimiter as /

input
=====
/home/smilyface/aaa/bbb
output
======
/home/smilyface/aaa