Error in my bash script (linux) counter is not counting during while statement

Question

im doing an assignment where i have to make a script, and it doesn't seem to be adding to the counter to display the clues when the counter reaches a certain number Here is the code below

#!/bin/bash
#This is a game that is played by the user having to guess a number that the computer thinks of.
counter=0
# functions
function clueOne() {
 echo “The number is over 50”
 return
}
function clueTwo() {
 echo “The number is smaller than 100 and is devided by 7, 10 times”
 return
}


function selectANumber() {
 dialog --backtitle "Number Game" --title "Number Game" --infobox "Thinking of a number..." 
10 50 ; clear
}
selectANumber
# game starts here
while [ $counter  -le 10 ];  do
 read -p ”Guess my number human!”
 if [ $counter -eq 5 ]; 
 then 
 clueOne
 counter=$((counter+1))
 elif [ $counter -eq 9 ]; 
 then
 clueTwo
 counter=$((counter+1))
 elif [ $REPLY = 70 ]; 
 echo “Yes that is my number!”; 
 exit 1
 echo “Wrong number”
 counter=$((counter+1))
fi
done
echo “Sorry you ran out of attempts”

Okay thank you very much, i have encountered another problem however, it's not displaying the clues when i get to a certain amount of attempts, would you know why? — Matt Scutt
– Matt Scutt, Commented Nov 20, 2014 at 7:10

anubhava · Accepted Answer · 2014-11-20 06:44:01Z

4

You're missing a then after this line:

elif [ $REPLY = 70 ];

Better run your script with http://www.shellcheck.net once to verify all the syntax.

answered Nov 20, 2014 at 6:44

anubhava

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