Exclude string from wildcard in bash
Posted
by Peter O'Doherty
on Stack Overflow
See other posts from Stack Overflow
or by Peter O'Doherty
Published on 2010-04-27T15:18:38Z
Indexed on
2010/04/27
15:33 UTC
Read the original article
Hit count: 426
bash
Hi, 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
© Stack Overflow or respective owner