increment variable in a string in shell script

问题

I have a string which comes from a variable I want to increment it. how can i do that using shell script?

this is my input which comes from a variable:

abc-ananya-01

output should be:

abc-ananya-02

回答1:

check this:

kent$  echo "abc-ananya-07"|awk -F'-' -v OFS='-' '{$3=sprintf("%02d",++$3)}7' 
abc-ananya-08

The above codes do the increment and keep your number format.

回答2:

It is shorter:

a=abc-lhg-08
echo ${a%-*}-`printf "%02d" $((10#${a##*-}+1))`
abc-lhg-09

Even better:

a=abc-lhg-08
printf "%s-%02d\n" ${a%-*} $((10#${a##*-}+1))
abc-lhg-09

回答3:

Bash string manipulations can be used here.

a='abc-ananya-07'
let last=$(echo ${a##*-}+1)
echo ${a%-*}-$(printf "%02d" $last)

回答4:

With pure Bash it is a bit long:

IFS="-" read -r -a arr <<< "abc-ananya-01"
last=10#${arr[${#arr}-1]}    # to prevent getting 08, when Bash 
                             # understands it is a number in base 8
last=$(( last + 1 ))
arr[${#arr}-1]=$(printf "%02d" $last)
( IFS="-"; echo "${arr[*]}" )

This reads into an array, increments the last element and prints it back.

It returns:

abc-ananya-02

来源：https://stackoverflow.com/questions/36933682/increment-variable-in-a-string-in-shell-script