Remove last character from line

Question

I want to remove last character from a line:

[root@ozzesh ~]#df -h | awk  '{ print $5 }'
Use%
22%
1%
1%
59%
51%
63%
5%

Expected result:

Use
22
1
1
59
51
63
5

Answer 1

sed 's/.$//'

To remove the last character.

But in this specific case, you could also do:

df -P | awk 'NR > 1 {print $5+0}'

With the arithmetic expression ($5+0) we force awk to interpret the 5th field as a number, and anything after the number will be ignored.

Note that GNU df (your -h is already a GNU extension, though not needed here) can also be told to only output the disk usage percentage:

df --output=pcent | tail -n +2 | tr -cd '0-9\n'

(tail skips the headers and tr removes everything but the digits and the line delimiters).

On Linux, see also:

findmnt -no USE%

Answer 2

With sed, this is pretty easy:

$ cat file
Use%
22%
1%
1%
59%
51%
63%
5%
$ sed 's/.$//' file
Use
22
1
1
59
51
63
5

The syntax is s(ubstitute)/search/replacestring/. The . indicates any character, and the $ the end of the line. So .$ will remove the last character only.

In this case, your complete command would look:

df -h | awk '{ print $5}' | sed 's/.$//'

Answer 3

I have two solutions :

1. cut: echo "somestring1" | rev | cut -c 2- | rev

   Here you reverse the string and cut the string from 2nd character and reverse again.

2. sed : echo "somestring1" | sed 's/.$//'

   Here you will search for regular expression .$ which means any characters followed by a last character and replace it with null // (between the two slashes)

Answer 4

Did you know that head and tail can work on a character basis instead of per line?

$ printf "I don't like periods." | head -c-1
I don't like periods

(BTW, printf is used here to prevent printing the "new line" character at the end of the line. If your output has a new line character and you want to remove that as well as the last non-whitespace character, use head -c-2).

This is basically the same as "Guru"'s solution using cut, but without the funky rev back and forth that they need to use because cut has no syntax for "last n things".

To iterate over a stream of lines and remove the last character of each, one might do something like this:

some-command | while read line; do echo $(head -c-2 <<<"$line"); done

Answer 5

In awk, you could do one of

awk '{sub(/%$/,"",$5); print $5}'
awk '{print substr($5, 1, length($5)-1)}'

Answer 6

echo "123" | perl -ple 'chop'
12

Answer 7

df -h | awk 'NR > 1{ print $5 }' | cut -d "%" -f1

Answer 8

another approach:

mapfile -t list < <(df -h)
printf '%s\n' "${list[@]%?}"

Turn it into a function:

remove_last() {
  local char=${1:-?}; shift
  mapfile -t list < <("$@")
  printf '%s\n' "${list[@]%$char}"
}

Then call it like this:

remove_last '%' df -h

mapfile is a bash4 feature.

The catch here is that you must supply a character to remove; if you want it to just be whatever the last character happens to be then you must pass '?' or ''. quotes required.

Answer 9

$ df -h | awk '{print $5}' | cut -d '%' -f1 

Answer 10

Try with this:

df -h | awk  '{ print $5 }' | sed "s/%//"

The normal use is: (ie)

VALUE=987654321
echo X123456789X | sed "s/123456789/${VALUE}/"

Response should be: X987654321X

Answer 11

sed -ie '$d' filename

here -i is to write changes
      e means expression
      $ means last line
      d means delete

Note:Without -e option $ wont work

Optional: To delete 1st and last line use sed -ie '1d;$d' filename

Answer 12

Using Raku (formerly known as Perl_6)

~$ printf '123\n456\n789' | raku -pe '.=chop;'
12
45
78

#OR:

~$ printf '123\n456\n789' | raku -ne '.chop.put;'
12
45
78

OR:

~$ printf '123\n456\n789' | raku -pe 's/.$//;'
12
45
78

#OR:

~$ printf '123\n456\n789' | raku -ne 'put S/.$//;'
12
45
78

Just like Perl, Raku can perform sed-like substitution using the -pe (autoprinting) command-line flags, or awk-like substitution using the -ne (non-autoprinting) flags. The Raku code above works fine even if the input text contains a blank line. Note the fourth example uses Raku's new S/// "big-S" non-destructive substitution notation, which "leaves the original string intact and returns the resultant string."

---

OP's specific example parsing the output of df -P:

~$ df -P | raku -ne 'put .words[4].subst("%");'

#OR:

~$ df -P | raku -ne 'put .words[4].trans("%" => "");'

To skip the first line (like the excellent answer from @StéphaneChazelas), use Raku's new (pre-incrementing) anonymous variable (++$):

~$ raku -ne  'if ++$ > 1 {put .words[4].subst("%") }'

#OR:

~$ raku -ne  'if ++$ > 1 {put .words[4].trans("%" => "") }'

https://docs.raku.org/language/5to6-nutshell
https://docs.raku.org/syntax/S%2F%2F%2F%20non-destructive%20substitution

Answer 13

df -h | awk {'print $5'}

Filesystem      Size  Used Avail Use% Mounted on
devtmpfs        4,0M     0  4,0M   0% /dev
tmpfs            16G  1,6G   15G  10% /dev/shm
tmpfs           6,3G  2,3M  6,3G   1% /run
efivarfs        148K   61K   83K  43% /sys/firmware/efi/efivars
/dev/nvme0n1p3  930G  874G   55G  95% /
/dev/nvme0n1p3  930G  874G   55G  95% /home
/dev/nvme0n1p2  974M  332M  576M  37% /boot
tmpfs            16G  135M   16G   1% /tmp
/dev/nvme0n1p1  599M   43M  556M   8% /boot/efi
tmpfs           3,2G  204K  3,2G   1% /run/user/1000
/dev/nvme0n1p3  930G  874G   55G  95% /swap

df -h | awk {'print $5'} | grep -v Use | sed 's|.$||g'

0
10
1
43
95
95
37
1
8
1
95

awk {'print $5'} = only prints the 5th word

grep -v Use = removes lines with the word 'Use'

sed 's|.$||g' = removes the last character