How to remove leading and trailing whitespaces?

Question

I'm using awk '{gsub(/^[ \t]+|[ \t]+$/,""); print;}' in.txt > out.txt to remove both leading and trailing whitespaces.

The problem is the output file actually has trailing whitespaces! All lines are of the same length - they are right padded with spaces.

What am I missing?

UPDATE 1

The problem is probably due to the the fact that the trailing spaces are nor "normal" spaces but \x20 characters (DC4).

UPDATE 2

I used gsub (/'[[:cntrl:]]|[[:space:]]|\x20/,"") an it worked. Two strange things:

1. Why isn't \x20 considered a control character?

2. Using '[[:cntrl:][:space:]\x20 does NOT work. Why?

Add

UPDATE: perhaps these are not simple spaces, but DC4 control characters? The files originated from Windows. – Add

\x20 is a regular ASCII space. Control characters are \x00 through \x1F. – Sick

Answer 1

This command works for me:

$ awk '{$1=$1}1' file.txt

Gereld

+1 Yes, why not? ;-) You can even do: awk '$1=$1' file.txt isn't it? – Dodder

@oHessling $1=$1 will delete empty line – Gereld

@eddi. awk will normalize a line by removing extra spaces. $1=$1 trigger the action, otherwise nothing will happen. – Gereld

I think you should add that to the answer together with an explanation of what the 1 does. – Josefinejoseito

@Josefinejoseito The 1 is the same as {print}. It will print every line. – Gereld

@Gereld : it does not work with GNU Awk 3.1.7 on CentOS 6.5 with ksh: echo "foo;bar ">tt && print "_$( awk -F";" -OFS";" '{$2=$2}1' tt)_" gives _foo;bar _. Did I miss something ? What is your setup, btw ? – Murillo

So awk '{$2=$2}1' is equivalent to awk '{$2=$2; print $0}'. For some awks; this doesn't strip whitespaces on all awks (e.g.Mac system awk). Also, you need to explain the quirky idiom in your answer. – Amalita

Answer 2

Your code is OK for me.
You may have something else than space and tabulation...
hexdump -C may help you to check what is wrong:

awk '{gsub(/^[ \t]+|[ \t]+$/,""); print;}' in.txt | hexdump -C | less

UPDATE:

OK you identified DC4 (there may be some other control characters...)
Then, you can improve your command:

awk '{gsub(/^[[:cntrl:][:space:]]+|[[:cntrl:][:space:]]+$/,""); print;}' in.txt > out.txt

See awk manpage:

[:alnum:] Alphanumeric characters.
[:alpha:] Alphabetic characters.
[:blank:] Space or tab characters.
[:cntrl:] Control characters.
[:digit:] Numeric characters.
[:graph:] Characters that are both printable and visible. (A space is printable, but not visible, while an a is both.)
[:lower:] Lower-case alphabetic characters.
[:print:] Printable characters (characters that are not control characters.)
[:punct:] Punctuation characters (characters that are not letter, digits, control characters, or space characters).
[:space:] Space characters (such as space, tab, and formfeed, to name a few).
[:upper:] Upper-case alphabetic characters.
[:xdigit:] Characters that are hexadecimal digits.

Leading/Trailing 0x20 removal

For me the command is OK, I have tested like this:

$ echo -e "\x20 \tTEXT\x20 \t" | hexdump -C
00000000  20 20 09 54 45 58 54 20  20 09 0a                 |  .TEXT  ..|
0000000b
$ echo -e "\x20 \tTEXT\x20 \t" | awk '{gsub(/^[[:cntrl:][:space:]]+|[[:cntrl:][:space:]]+$/,""); print;}' | hexdump -C
00000000  54 45 58 54 0a                                    |TEXT.|
00000005

However if you have 0x20 in the middle of your text
=> then it is not removed.
But this is not your question, isn't it?

Dodder

I really thought this would work, but it didn't, I'm still left with all these "spaces" ASCII code 20 (int = 32). – Add

Hello @user1194552. Please provide your hexdump -C output before and after awk processing. Then I can better understand your issue. Because when I try to test, it looks good for me :-) – Dodder

What is your awk --version? I can test two versions: GNU Awk 3.1.3 and GNU Awk 3.1.5. And please provide your hexdump -C. Then I can test the same thing as you. – Dodder

Answer 3

Your files probably have Windows line endings. That means that they end with \r\n, so matching a sequence of tabs and spaces at the end of the line won't work -- awk tries to match all the tabs and spaces that come after the \r. Try running the file through tr -d "\r" before sending it to awk.

Innocency

Answer 4

Perl could be used:

perl -lpe 's/^\s*(.*\S)\s*$/$1/' in.txt > out.txt

s/foo/bar/ substitute using regular expressions
^ beginning of string
\s* zero or more spaces
(.*\S) any characters ending with a non-whitespace. Capture it into $1
\s* zero or more spaces
$ end of string

Bastien