How to pull a string apart by its contents

Question

I have a string in a common pattern that I want to manipulate. I want to be able to turn string 5B299 into 5B300 (increment the last number by one).

I want to avoid blindly splicing the string by index, as the first number and letter can change in size. Essentially I want to be able to get the entire value of everything after the first character, increment it by one, and re-append it.

The only things I've found online so far show me how to cut by a delimiter, but I don't have a constant delimiter.

Answer 1

You could use the regex features supported by the bash shell with its ~ construct that supports basic Extended Regular Expression matching (ERE). All you need to do is define a regex and work on the captured groups to get the resulting string

str=5B299
re='^(.*[A-Z])([0-9]+)$' 

Now use the ~ operator to do the regex match. The ~ operator populates an array BASH_REMATCH with the captured groups if regex match was successful. The first part (5B in the example) would be stored in the index 0 and the next one at 1. We increment the value at index 1 with the $((..)) operator.

if [[ $str =~ $re ]]; then
    result="${BASH_REMATCH[1]}$(( BASH_REMATCH[2] + 1 ))"
    printf '%s\n' "$result"
fi

The POSIX version of the regex, free of the locale dependency would be to use character classes instead of range expressions as

posix_re='^(.*[[:alpha:]])([[:digit:]]+)$'

Answer 2

You can do what you are attempting fairly easily with the bash parameter-expansion for string indexes along with the POSIX arithmetic operator. For instance you could do:

#!/bin/bash

[ -z "$1" ] && {    ## validate at least 1 argument provided
    printf "error: please provide a number.\n" >&2
    exit 1
}

[[ $1 =~ [^0-9][^0-9]* ]] && {  ## validate all digits in argument
    printf "error: input contains non-digit characters.\n" >&2
    exit 1
}

suffix=${1:1}           ## take all character past 1st as suffix
suffix=$((suffix + 1))  ## increment suffix by 1

result=${1:0:1}$suffix  ## append suffent to orginal 1st character
echo "$result"          ## output

exit 0

Which will leave the 1st character alone while incrementing the remaining characters by 1 and then joining again with the original 1st digit, while validating that the input consisted only of digits, e.g.

Example Use/Output

$ bash prefixsuffix.sh
error: please provide a number.

$ bash prefixsuffix.sh 38a900
error: input contains non-digit characters.

$ bash prefixsuffix.sh 38900
38901

$ bash prefixsuffix.sh 39999
310000

Look things over and let me know if that is what you intended.

Answer 3

You can use sed in conjunction with awk:

increment() {
   echo $1 | sed -r 's/([0-9]+[a-zA-Z]+)([0-9]+)/\1 \2/' | awk '{printf "%s%d", $1, ++$2}'
}

echo $(increment "5B299")
echo $(increment "127ABC385")
echo $(increment "7cf999")

Output:

5B300
127ABC386
7cf1000