Shell script doesn't run

Question

Can someone help and tell me why this isn't working?

I have checked the script and solved some problems but it still doesn't works fine and I can't find the mistake.

#!/bin/bash

#variabelen

IP="192.168.0."
array=($@)

#functies

What's wrong with the array? My linux told me that there is a syntax error on line 12 so with that array but he doesn't tell me what.

function Sorteer(){

array=($(printf '%s\n' "$@"|sort -nu))
for i in ${array[@]};do
ping -c -1 "IP"$i
done

}

function Telbij(){
# given number $i +200
    b=$(( $i + 200 ))    
    if (( b > 255 ))
    then
        echo "Neem kleiner getal";  
    else
    ping -c 1 "IP"$b;
    fi
}

function XXYY() {
#ping 65-68 = ping 65 66 67 68
start=${1%-*}
end=${1#*-}
for ((i=start;i<=end;i++));do
    ping -c 1 "$IP"$i
done
}

The mistake is in the if else function: http://prntscr.com/7gr8yf

But I don't know what that means: "The mentioned parser error was in this else clause."

if [ "$#" -eq 0 ]; then   
        echo "Er moet minimaal 1 parameter worden meegegeven "
        exit 0                 
   else

case

-h -help ) echo "Geef de laatste cijfers van het IP-adres van de pc's om te testen.";;

XX-YY ) XXYY;;

-t ) Telbij;;

- sort ) Sorteer;;

esac

fi
done

Answer 1

Don't know what's specifically not working but,

in Sorteer function you should double quote array expansions to avoid re-splitting elements. Try change to following:

function Sorteer(){

array=($(printf '%s\n' "$@"|sort -nu))
for i in "${array[@]}";do
ping -c -1 "IP"$i
done

}

Now your case operator should be like:

case $some_value in

-help ) echo "Geef de laatste cijfers van het IP-adres van de pc's om te testen.";;

XX-YY ) XXYY;;

-t ) Telbij;;

-sort ) Sorteer;;

esac

This will fix your if issue as well