I am trying to write a bash script that takes in an option. Lets call these options A and B.
In the script A and B may or may not be defined as variables.
I want to be able to check if the variable is defined or not.
I have tried the following but it doesn't work.
if [ ! -n $1 ]; then
echo "Error"
fi
Thanks
The "correct" way to test whether a variable is set is to use the + expansion option. You'll see this a lot in configure scripts:
if test -s "${foo+set}"
where ${foo+set} expands to "set" if it is set or "" if it's not. This allows for the variable to be set but empty, if you need it. ${foo:+set} additionally requires $foo to not be empty.
(That $(eval echo $a) thing has problems: it's slow, and it's vulnerable to code injection (!).)
Oh, and if you just want to throw an error if something required isn't set, you can just refer to the variable as ${foo:?} (leave off the : if set but empty is permissible), or for a custom error message ${foo:?Please specify a foo.}.
-v test, as mentioned in my answer, to denote if a variable is set.You did not define how these options should be passed in, but I think:
if [ -z "$1" ]; then
echo "Error"
exit 1
fi
is what you are looking for.
However, if some of these options are, err, optional, then you might want something like:
#!/bin/bash
USAGE="$0: [-a] [--alpha] [-b type] [--beta file] [-g|--gamma] args..."
ARGS=`POSIXLY_CORRECT=1 getopt -n "$0" -s bash -o ab:g -l alpha,beta:,gamma -- "$@"`
if [ $? -ne 0 ]
then
echo "$USAGE" >&2
exit 1
fi
eval set -- "$ARGS"
unset ARGS
while true
do
case "$1" in
-a) echo "Option a"; shift;;
--alpha) echo "Option alpha"; shift;;
-b) echo "Option b, arg '$2'"; shift 2;;
--beta) echo "Option beta, arg '$2'"; shift 2;;
-g|--gamma) echo "Option g or gamma"; shift;;
--) shift ; break ;;
*) echo "Internal error!" ; exit 1 ;;
esac
done
echo Remaining args
for arg in "$@"
do
echo '--> '"\`$arg'"
done
exit 0
Don't do it that way, try this:
if [[ -z $1 ]]; then
echo "Error"
fi
The error in your version is actually the lack of quoting.
Should be:
if [ ! -n "$1" ]; then
echo "Error"
fi
But you don't need the negation, use -z instead.
If you work on Bash, then use double brackets [[ ]] too.
from the man bash page:
-z string True if the length of string is zero. -n string True if the length of string is non-zero.
Also, if you use bash v4 or greater (bash --version) there's -v
-v varname True if the shell variable varname is set (has been assigned a value).
The trick is "$1", i.e.
root@root:~# cat auto.sh
Usage () {
echo "error"
}
if [ ! -n $1 ];then
Usage
exit 1
fi
root@root:~# bash auto.sh
root@root:~# cat auto2.sh
Usage () {
echo "error"
}
if [ ! -n "$1" ];then
Usage
exit 1
fi
root@root:~# bash auto2.sh
error
