how to find the last grouped digit in a string in bash - bash

Question

This is a follow-up question to this question, regarding how to know the number of grouped digits in string.
In bash,
How can I find the last occurrence of a group of digits in a string?
So, if I have
string="123 abc 456"
I would get
456
And if I had
string="123 123 456"
I would still get
456

Answer 1

Without external utilities (such as sed, awk, ...):
$ s="123 abc 456"
$ [[ $s =~ ([0-9]+)[^0-9]*$ ]] && echo "${BASH_REMATCH[1]}"
456
BASH_REMATCH is a special array where the matches from [[ ... =~ ... ]] are assigned to.
Test code:
str=("123 abc 456" "123 123 456" "123 456 abc def" "123 abc" "abc 123" "123abc456def")
for s in "${str[#]}"; do
[[ $s =~ ([0-9]+)[^0-9]*$ ]] && echo "$s -> ${BASH_REMATCH[1]}"
done
Output:
123 abc 456 -> 456
123 123 456 -> 456
123 456 abc def -> 456
123 abc -> 123
abc 123 -> 123
123abc456def -> 456

Answer 2

You can use a regex in Bash:
$ echo "$string"
123 abc 456
$ [[ $string =~ (^.*[ ]+|^)([[:digit:]]+) ]] && echo "${BASH_REMATCH[2]}"
456
If you want to capture undelimited strings like 456 or abc123def456 you can do:
$ echo "$string"
test456text
$ [[ $string =~ ([[:digit:]]+)[^[:digit:]]*$ ]] && echo "${BASH_REMATCH[1]}"
456
But if you are going to use an external tool, use awk.
Here is a demo of Bash vs Awk to get the last field of digits in a string. These are for digits with ' ' delimiters or at the end or start of a string.
Given:
$ cat file
456
123 abc 456
123 123 456
abc 456
456 abc
123 456 foo bar
abc123def456
Here is a test script:
while IFS= read -r line || [[ -n $line ]]; do
bv=""
av=""
[[ $line =~ (^.*[ ]+|^)([[:digit:]]+) ]] && bv="${BASH_REMATCH[2]}"
av=$(awk '{for (i=1;i<=NF;i++) if (match($i, /^[[:digit:]]+$/)) last=$i; print last}' <<< "$line")
printf "line=%22s bash=\"%s\" awk=\"%s\"\n" "\"$line\"" "$bv" "$av"
done <file
Prints:
line= "456" bash="456" awk="456"
line= "123 abc 456" bash="456" awk="456"
line= "123 123 456" bash="456" awk="456"
line= "abc 456" bash="456" awk="456"
line= "456 abc" bash="456" awk="456"
line= "123 456 foo bar" bash="456" awk="456"
line= "abc123def456" bash="" awk=""

Answer 3

grep -o '[0-9]\+' file|tail -1
grep -o lists matched text only
tail -1 output only the last match
well, if you have string:
grep -o '[0-9]\+' <<< '123 foo 456 bar' |tail -1

Answer 4

You may use this sed to extract last number in a line:
sed -E 's/(.*[^0-9]|^)([0-9]+).*/\2/'
Examples:
sed -E 's/(.*[^0-9]|^)([0-9]+).*/\2/' <<< '123 abc 456'
456
sed -E 's/(.*[^0-9]|^)([0-9]+).*/\2/' <<< '123 456 foo bar'
456
sed -E 's/(.*[^0-9]|^)([0-9]+).*/\2/' <<< '123 123 456'
456
sed -E 's/(.*[^0-9]|^)([0-9]+).*/\2/' <<< '123 x'
123
RegEx Details:
(.*[^0-9]|^): Match 0 or more characters at start followed by a non-digit OR line start.
([0-9]+): Match 1+ digits and capture in group #2
.*: Match remaining characters till end of line
\2: Replace it with back-reference #2 (what we captured in group #2)

Answer 5

Another way to do it with pure Bash:
shopt -s extglob # enable extended globbing - for *(...)
tmp=${string%%*([^0-9])} # remove non-digits at the end
last_digits=${tmp##*[^0-9]} # remove everything up to the last non-digit
printf '%s\n' "$last_digits"

Answer 6

This is a good job for parameter expansion:
$ string="123 abc 456"
$ echo ${string##* }
456

Answer 7

A simple answer with gawk:
echo "$string" | gawk -v RS=" " '/^[[:digit:]]+$/ { N = $0 } ; END { print N }'
With RS=" ", we read each field as a separate record.
Then we keep the last number found and print it.
$ string="123 abc 456 abc"
$ echo "$string" | gawk -v RS=" " '/^[[:digit:]]+$/ { N = $0 } ; END { print N }'
456