How to split a string in shell and get the last field

Question

Suppose I have the string 1:2:3:4:5 and I want to get its last field (5 in this case). How do I do that using Bash? I tried cut, but I

Answer 1

It's difficult to get the last field using cut, but here are some solutions in awk and perl

echo 1:2:3:4:5 | awk -F: '{print $NF}'
echo 1:2:3:4:5 | perl -F: -wane 'print $F[-1]'

Answer 2

You can use string operators:

$ foo=1:2:3:4:5
$ echo ${foo##*:}
5

This trims everything from the front until a ':', greedily.

${foo  <-- from variable foo
  ##   <-- greedy front trim
  *    <-- matches anything
  :    <-- until the last ':'
 }

Answer 3

Using Bash.

$ var1="1:2:3:4:0"
$ IFS=":"
$ set -- $var1
$ eval echo  \$${#}
0

Answer 4

Regex matching in sed is greedy (always goes to the last occurrence), which you can use to your advantage here:

$ foo=1:2:3:4:5
$ echo ${foo} | sed "s/.*://"
5

Answer 5

Another way is to reverse before and after cut:

$ echo ab:cd:ef | rev | cut -d: -f1 | rev
ef

This makes it very easy to get the last but one field, or any range of fields numbered from the end.

Answer 6

Assuming fairly simple usage (no escaping of the delimiter, for example), you can use grep:

$ echo "1:2:3:4:5" | grep -oE "[^:]+$"
5

Breakdown - find all the characters not the delimiter ([^:]) at the end of the line ($). -o only prints the matching part.