remove particular characters from a variable using bash
I want to parse a variable (in my case it’s development kit version) to make it dot(.) free. If version='2.3.3', desired output is 233.
I tried as below, but it requires . to be replaced with another character giving me 2_3_3. It would have been fine if tr . '' would have worked.
1 VERSION='2.3.3'
2 echo "2.3.3" | tr . _
You should try with sed instead
sed 's/.//g'
There is no need to execute an external program. bash‘s string manipulation can handle it (also available in ksh93 (where it comes from), zsh and recent versions of mksh, yash and busybox sh (at least)):
$ VERSION='2.3.3'
$ echo "${VERSION//.}"
233
(In those shells’ manuals you can generally find this in the parameter expansion section.)
By chronological order:
tr/sed
echo "$VERSION" | tr -d .
echo "$VERSION" | sed 's/.//g'
csh/tcsh
echo $VERSION:as/.//
POSIX shells:
set -f
IFS=.
set -- $VERSION
IFS=
echo "$*"
ksh93/zsh/mksh/bash/yash (and busybox ash when built with ASH_BASH_COMPAT)
echo "${VERSION//.}"
zsh
echo $VERSION:gs/./
In addition to the successful answers already exists. Same thing can be achieved with tr, with the --delete option.
echo "2.3.3" | tr --delete .
echo "2.3.3" | tr -d . # for MacOS
Which will output: 233
Perl
$ VERSION='2.3.3'
$ perl -pe 's/.//g' <<< "$VERSION"
233
Python
$ VERSION='2.3.3'
$ python -c 'import sys;print sys.argv[1].replace(".","")' "$VERSION"
233
If $VERSION only contains digits and dots, we can do something even shorter:
$ python -c 'print "'$VERSION'".replace(".","")'
233
(beware it’s a code injection vulnerability though if $VERSION may contain any character).
AWK
$ VERSION='2.3.3'
$ awk 'BEGIN{gsub(/./,"",ARGV[1]);print ARGV[1]}' "$VERSION"
233
Or this:
$ awk '{gsub(/./,"")}1' <<< "$VERSION"
233
echo "$VERSION" | tr -cd [:digit:]
That would return only digits, no matter what other characters are present