I'd like to be able to accept both mandatory and optional flags in my script. Here's what I have so far.
#!bin/bash
while getopts ":a:b:cdef" opt; do
case $opt in
a ) APPLE="$OPTARG";;
b ) BANANA="$OPTARG";;
c ) CHERRY="$OPTARG";;
d ) DFRUIT="$OPTARG";;
e ) EGGPLANT="$OPTARG";;
f ) FIG="$OPTARG";;
?) echo "Invalid option: -"$OPTARG"" >&2
exit 1;;
: ) echo "Option -"$OPTARG" requires an argument." >&2
exit 1;;
esac
done
echo "Apple is "$APPLE""
echo "Banana is "$BANANA""
echo "Cherry is "$CHERRY""
echo "Dfruit is "$DFRUIT""
echo "Eggplant is "$EGGPLANT""
echo "Fig is "$FIG""
However, the output for the following:
bash script.sh -a apple -b banana -c cherry -d dfruit -e eggplant -f fig
...outputs this:
Apple is apple
Banana is banana
Cherry is
Dfruit is
Eggplant is
Fig is
As you can see, the optional flags are not pulling the arguments with $OPTARG as it does with the required flags. Is there a way to read $OPTARG on optional flags without getting rid of the neat ":)" error handling?
=======================================
EDIT: I wound up following the advice of Gilbert below. Here's what I did:
#!/bin/bash
if [[ "$1" =~ ^((-{1,2})([Hh]$|[Hh][Ee][Ll][Pp])|)$ ]]; then
print_usage; exit 1
else
while [[ $# -gt 0 ]]; do
opt="$1"
shift;
current_arg="$1"
if [[ "$current_arg" =~ ^-{1,2}.* ]]; then
echo "WARNING: You may have left an argument blank. Double check your command."
fi
case "$opt" in
"-a"|"--apple" ) APPLE="$1"; shift;;
"-b"|"--banana" ) BANANA="$1"; shift;;
"-c"|"--cherry" ) CHERRY="$1"; shift;;
"-d"|"--dfruit" ) DFRUIT="$1"; shift;;
"-e"|"--eggplant" ) EGGPLANT="$1"; shift;;
"-f"|"--fig" ) FIG="$1"; shift;;
* ) echo "ERROR: Invalid option: ""$opt""" >&2
exit 1;;
esac
done
fi
if [[ "$APPLE" == "" || "$BANANA" == "" ]]; then
echo "ERROR: Options -a and -b require arguments." >&2
exit 1
fi
Thanks so much, everyone. This works perfectly so far.
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