Removing final bash script argument
- by ctuffli
I'm trying to write a script that searches a directory for files and greps for a pattern. Something similar to the below except the find expression is much more complicated (excludes particular directories and files).
#!/bin/bash
if [ -d "${!#}" ]
then
path=${!#}
else
path="."
fi
find $path -print0 | xargs -0 grep "$@"
Obviously, the above doesn't work because "$@" still contains the path. I've tried variants of building up an argument list by iterating over all the arguments to exclude path such as
args=${@%$path}
find $path -print0 | xargs -0 grep "$path"
or
whitespace="[[:space:]]"
args=""
for i in "${@%$path}"
do
# handle the NULL case
if [ ! "$i" ]
then
continue
# quote any arguments containing white-space
elif [[ $i =~ $whitespace ]]
then
args="$args \"$i\""
else
args="$args $i"
fi
done
find $path -print0 | xargs -0 grep --color "$args"
but these fail with quoted input. For example,
# ./find.sh -i "some quoted string"
grep: quoted: No such file or directory
grep: string: No such file or directory
Note that if $@ doesn't contain the path, the first script does do what I want.
