CODEWITHSUNDEEP
bash
J Freebird
J Freebird

Reputation: 3910

Bash Script No Such File or Directory when Resolving Passed-in Arguments

I have a bash script that looks like this:

if [ $# -eq 2 ]; then
    REDIS_HOME=$2
else
    if [ $# -eq 3 ]; then
        REDIS_HOME=$2
        PORTS=$3
    else
        REDIS_HOME="/usr/local/redis"
    fi
fi

case "$1" in
    start)
       if [ -z "$PORTS" ]; then
           cmd="$REDIS_HOME/bin/redis-server $REDIS_HOME/redis-6379.conf"
           $cmd
       else
           IFS=",";
           for port in $PORTS
           do
               cmd="$REDIS_HOME/bin/redis-server $REDIS_HOME/redis-$port.conf"
               $cmd
           done
       fi
    ;;
    *)
    ;;
esac

exit 0

When I run the script with my_script.sh start, it works well by using the default redis home in the script. But when I ran it with my_script.sh start /usr/local/redis 6379, it says "/usr/local/redis/bin/redis-server /usr/local/redis/redis-6379.conf" No such file or directory.

Basically, I'm passing in the same REDIS_HOME, but I cannot figure out why the script cannot resolve the path if it's passed in as a parameter.

Upvotes: 0

Views: 1180

Answers (1)

Barmar
Barmar

Reputation: 781848

IFS=, is causing the problem. When you do

$cmd

it uses $IFS to break the expansion of $cmd into words. Since space is not in $IFS, the space is not treated as a delimiter between the program name and the argument, so the entire result is treated as the program name. And of course it's not found.

I'm not sure why you need the $cmd variable in the first place. You can just do:

IFS=",";
for port in $PORTS
do
    $REDIS_HOME/bin/redis-server $REDIS_HOME/redis-$port.conf
done

Upvotes: 2

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