Exclude string from wildcard in bash - bash

Question

I'm trying to adapt a bash script from "Sams' Teach Yourself Linux in 24 Hours" which is a safe delete command called rmv. The files are removed by calling rmv -d file1 file2 etc. In the original script a max of 4 files can by removed using the variables $1 $2 $3 $4.
I want to extend this to an unlimited number of files by using a wildcard.
So I do:
for i in $*
do
mv $i $HOME/.trash
done
The files are deleted okay but the option -d of the command rmv -d is also treated as an argument and bash objects that it cannot be found. Is there a better way to do this?
Thanks,
Peter
#!/bin/bash
# rmv - a safe delete program
# uses a trash directory under your home directory
mkdir $HOME/.trash 2>/dev/null
# four internal script variables are defined
cmdlnopts=false
delete=false
empty=false
list=false
# uses getopts command to look at command line for any options
while getopts "dehl" cmdlnopts; do
case "$cmdlnopts" in
d ) /bin/echo "deleting: \c" $2 $3 $4 $5 ; delete=true ;;
e ) /bin/echo "emptying the trash..." ; empty=true ;;
h ) /bin/echo "safe file delete v1.0"
/bin/echo "rmv -d[elete] -e[mpty] -h[elp] -l[ist] file1-4" ;;
l ) /bin/echo "your .trash directory contains:" ; list=true ;;
esac
done
if [ $delete = true ]
then
for i in $*
do
mv $i $HOME/.trash
done
/bin/echo "rmv finished."
fi
if [ $empty = true ]
then
/bin/echo "empty the trash? \c"
read answer
case "$answer" in
y) rm -i $HOME/.trash/* ;;
n) /bin/echo "trashcan delete aborted." ;;
esac
fi
if [ $list = true ]
then
ls -l $HOME/.trash
fi

Answer 1

You can make use of shift here.
Once you find -d is one of the options in the switch, you can shift and get rid of -d from the positional parameters. Next you can
until [ -z $1 ] ; do
mv $1 $HOME/.trash
shift
done

Answer 2

getopts sets OPTIND to the index of the first argument after the options. (#)
So after parsing the options you can do:
shift $OPTIND-1
to remove the options from the argument list.
Then use "$#" rather than $*, and you can handle files with spaces in them.

Answer 3

Thanks a lot!
I changed the code to read:
#!/bin/bash
# rmv - a safe delete program
# todo: add ability to handle wildcards
# uses a trash directory under your home directory
mkdir $HOME/.trash 2>/dev/null
# four internal script variables are defined
cmdlnopts=false
delete=false
empty=false
list=false
# uses getopts command to look at command line for any options
while getopts "dehl" cmdlnopts; do
case "$cmdlnopts" in
d ) echo -e "deleting: \n" "${#:2}" ; delete=true ;;
e ) echo -e "emptying the trash..." ; empty=true ;;
h ) echo -e "safe file delete v1.0"
echo -e "rmv -d[elete] -e[mpty] -h[elp] -l[ist] file [...]" ;;
l ) echo -e "your .trash directory contains:" ; list=true ;;
esac
done
shift $OPTIND-1
if [ $delete = true ]
then
for i in $#
do
mv $i $HOME/.trash
done
echo "rmv finished."
fi
then
/bin/echo "empty the trash? \c"
read answer
case "$answer" in
y) rm -i $HOME/.trash/* ;;
n) /bin/echo "trashcan delete aborted." ;;
esac
fi
if [ $list = true ]
then
ls -l $HOME/.trash
fi
This deletes the files as desired but I get this error:
/home/peter/rmv: line 21: shift: 2-1: numeric argument required
mv: invalid option -- 'd'
Try `mv --help' for more information.

Answer 4

You need to use
shift $(($OPTIND - 1))
to get red of the processed command line args. Try this version:
#!/bin/bash
# rmv - a safe delete program
# uses a trash directory under your home directory
mkdir -p $HOME/.trash
# uses getopts command to look at command line for any options
while getopts "dehl" cmdlnopts; do
case "$cmdlnopts" in
d ) delete=true;;
e ) echo "emptying the trash..." ; empty=true ;;
h ) echo "safe file delete v1.0"
echo "rmv -d[elete] -e[mpty] -h[elp] -l[ist] files" ;;
l ) echo "your .trash directory contains:" ; list=true ;;
esac
done
shift $(($OPTIND - 1))
if [ -n "${delete}" ]; then
echo "deleting: " "${#}"
mv ${#} $HOME/.trash
echo "rmv finished."
fi
if [ -n "${empty}" ]; then
read -p "empty the trash? " answer
case "$answer" in
y) rm -i $HOME/.trash/* ;;
n) echo "trashcan delete aborted." ;;
esac
fi
if [ -n "${list}" ]; then
ls -l $HOME/.trash
fi

Answer 5

Late to the party, but for Googlers, this will generate the error Peter describes:
shift $OPTIND-1
while the syntax in Jurgen's reply will not:
shift $(($OPTIND-1))
The problem is that $OPTIND-1 is interpreted as a string, and shift can't use a string as an argument. $(( )) is Bash's arithmetic expansion operator. You put a string inside it, the string is evaluated as an arithmetic expression, and the value returned.