how to get rid of trailing white spaces using awk

[Kent Ostby]

Fernando -- I'm making the assumption that you only want to get rid of white space at the end of the line.

sample data input (add three spaces after last word):

Fernando Fulgar Fernando

To test:
wc indata
1 3 30
{NOTE: one character is a carriage return)

Create a file, tw.awk, with this script in it:

{dalen=length($0);daflag=0;
for (idx1=dalen;daflag==0;idx1--)
{ if (substr($0,idx1,1)!=" ")
{print substr($0,1,idx1);exit}
}
}

awk -f tw.awk < indata > outdata

wc -l outdata
1 3 27 outdata

best regards,

Kent M. Ostby

[Hein van den Heuvel]

I like perl a lot, but in this case there is a perfectly reasonable, compact, readable awkward solution as requested.

As I replied earlier, but I guess is got overlooked, just use the SUBstitute function using "any number of spaces at end of variable" as ere (extended regular expression) and nothing ("") as replacement string on the variable of choice.
This case $2, and often you can just work on $0.

Solution:

sub(/ *$/,"",$2)


Cheers,
Hein.

[Alexandre Dumont_1]

If possible, i would rather use sed to get rid of heading and trailing spaces:

sed -e "s/^[ \t]*//; s/[ \t]*$//;"

Regards,
Alex

[Pando]

dear all,

it was a great help!
many thanks!