how to get rid of trailing white spaces using awk

[Pando]

how can i remove of the traing white spaces using awk.

I have used the command below to get the value but i also get the white spaces with it.

mode=$(awk -F "," '/Test Mode/ {print $2}' $MFILE)

Maximum points for al correct replies.

[H.Merijn Brand (procura]

does it /have/ to be awk?

# perl -ple's/\s+$//' file

And to strip trailing whitespace in-place (the file itself will actually be changed):

# perl -pli -e's/\s+$//' file

Enjoy, Have FUN! H.Merijn

[Gordon Morrison_1]

Why be AWKward? ksh can do it.
If it is a single "word" with training spaces, declare the variable that will hold it with:
typeset -L varname
or
shortline=${line%%[ ]*}
(Put a tab and a space inside the []'s

If it's a line with more than one word on it, pass the line variable to a function, then extract each word with $1 $2 $3 etc. within that function (Don't pass the variable inside quotes, or it will take the whole thing to be $1)

[Peter Godron]

Fernando,
with sed:
to delete whitespaces from end of line
sed 's/[ \t]*$//'
to delete whitespaces from start and end of line
sed 's/^[ \t]*//;s/[ \t]*$//'

\t represents the tab key!

Regards

[Leif Halvarsson_2]

Hi,
use the gsub function in awk.

gsub(" ","",$MFILE); print $MFILE

[Noel Miranda]

You can also pipe the output to awk again, this time printing the first field, with separator being the default, i.e white space.

[Hein van den Heuvel]

untested:

mode=$(awk -F "," '/Test Mode/ { sub(/ *$/,"",$2); print $2}' $MFILE)


The ere specifies any number of spaces at the end of $2 to be replaced with nothing.

Hein.

[Amit Agarwal_1]

you can try combination of match and substr.

mode=$(awk -F "," '/Test Mode/ {print substr($2, 1, match($2, "[ \t]*$")-1)}' $MFILE)


-Amit

[Ionut Grigorescu_2]

Fernando, you should give some points to Noel, so why not pipe it to awk again, like:
awk -F "," '/Test Mode/{print $2}' $MFILE|awk -F " " '{print $1}'
..but still I would use perl...

[Amit Agarwal_1]

Fernando is using "," as FS, so I tend to believe that each field can be in form of multiple words. *IF* it is so, then Noel's solution won't work.

[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!