Bash Programming how to call "$i+1" - linux

Question

I am writing a script that will allow me to change a char in a string from "#" to something else, if I call an argument in terminal.
eg if I write
./myprogram testText.txt -r a
the -r argument will remove all "#" from testTxt.txt and replace them with "a"
My problem is I do not know how to write "If -r is $x, $x+1 is the char I want for replacement"
This is purely a syntax problem, I'm a bash noob :P. Here is the part of code I'm trying to work with.
for i in $*
do
if [[ $i = "-r" ]]
then
$customHashChoice=$((i+1))
# ^^^^^ Problematic Line ^^^^
fi
done

Answer 1

Try this:
customHashChoice=($(getopt "r:" "$#" 2>/dev/null))
if [ "${customHashChoice[0]}" == "-r" ]; then
customHashChoice="${customHashChoice[1]}"
else
echo "-r option is missing. Aborting..."
exit 1
fi
Syntax: getopt optstring parameters
From manual: getopt is used to break up (parse) options in command lines for easy parsing by shell procedures, and to check for legal options. It uses the GNU getopt(3) routines to do this.
Here, optstring is r:. It means, that the script accepts an option -r & the option takes an argument (implied by :).
The output of getopt "r:" "$#" is as below:
-r <argument to -r option> -- <unmatched parameters>
e.g. for command-line arguments,
./myprogram testText.txt -r a
getopt "r:" "$#" returns
-r a -- testText.txt
This output is stored in array & the second element of array is used, if the first element is equal to -r.

Answer 2

i=1
while [ "$i" -le $# ]
do
if [[ ${!i} = "-r" ]]
then
i=$(($i + 1))
customHashChoice=${!i}
i=$(($i + 1))
continue
fi
# do something useful
i=$(($i + 1))
done
The command line arguments are numbered 1 through $#. The above loops through each of them. If first checks if the current argument is -r and, if so, sets customHashChoice.
In the above, i contains the argument number. So, $i gives the value of i. To access the i'th command line argument, one uses ${!i}.
A more standard approach
The standard way to process command line arguments in shell scripts is getopts. It can handle many options. Here is sample code that that takes an option -r and requires it to have an argument, which is assigned to the shell variable char:
while getopts r: arg ; do case $arg in
r) char="$OPTARG" ;;
:) echo "${0##*/}: Must supply an argument to $OPTARG." ; exit 1 ;;
\?) echo "Invalid option" ; exit 1 ;;
esac
done
shift $(($OPTIND - 1))
echo "I will replace # with $char in file $1"
For getopts to work, the options have to come first. So, your command line would becomes:
./myprogram -r a testText.txt
If this is not acceptable, you can roll your own custom option processor. In the long run, there is some advantage, however, to standardizing on the usual approach.

Answer 3

You could do something like the following:
#!/bin/bash
val=
xval=
fname=$1
while [ "$*" != "" ]; do
case $1 in
"-r") val="${2}"; shift ;;
"-x") xval="${2}"; shift ;;
esac
shift
done
echo ${fname} ${val} ${xval}
Then when you pass the command like so
./myprogram testText.txt -r a
fname will be testText.txt, and the arguments will be parsed (where the -r will pick up a); for any other values you might want to parse, you'll need variable names to assign and test against. The output would be:
testText.txt a
Hope that helps