How to get the last digits of a string e.g print 201 for string ua07app201?
#sed back reference: print the first match pattern enclosed by ( )
echo ua07app201 | sed 's/.*[^0-9]\([0-9]*\)$/\1/'
#Sed delete: Delete the longest match of non-digits char
echo 'ua07app201' | sed 's/.*[^0-9]//'
#Expr matching operator : Similar to sed back reference, without (), it returns the the number of matched chars.
expr ua07app201 : '.*[^0-9]\([0-9]*\)$'
#Awk: set non-digit as seperator, print the last filed $NF
echo 'ua07app201' | nawk -F'[^0-9]' '{print $NF}'
#Perl in command line mode
echo ua07app201 | perl -nle ' $_ =~m /(\d+$)/; print $1'
or the simplizied version
echo ua07app201|perl -nle'print/(\d+)$/'
#Parameter Substitution, delete the longest match of non-digits chars from beginning.
a='ua07app201';echo "${a##*[!0-9]}"
How to get path only from full path of a file?
#Parameter Substitution, delete the shortest match from end
$ var=/var/tmp/test.txt;echo ${var%/*}
/var/tmp
How to sort a string?
#one-liner to sort a string
$echo "s03 s08 s01" | tr '[:space:]' '\n' | sort -n | paste -s
s01 s03 s08
Posts
