CODEWITHSUNDEEP
bash
Webtm
Webtm

Reputation: 87

Counter in BASH to count the times a number was entered?

I am trying to write a script where you enter a number, then enter another value, if that value is the same as the first number entered, it adds to the cnt variable, then lets you input another number to check. It is supposed to repeated this until the two numbers do not match, and then print how many times the value occurred. I'm having trouble with the loop, after the second value is entered, it prints continuously: 5 occurs 1 times

    #!/bin/bash
read currVal
if [ -n $currVal ]; then
    cnt=1
    read val
    while [[ -n $val ]]
    do
            if [[ $val == $currVal ]]; then
                    cnt=$((cnt+1))
            else
                    echo "$currVal occurs $cnt times"
            fi
    done

fi

Upvotes: 0

Views: 246

Answers (2)

clt60
clt60

Reputation: 63972

A variation what counts the entered values:

declare -A counts
while :
do
    read -r -p 'Enter number (or press enter to finish) > ' num
    [[ -z "$num" ]] && break
    ((counts["$num"]++))
done

for key in "${!counts[@]}"
do
    printf "entered %s: %s times\n" "$key" ${counts["$key"]}
done | sort -n

Upvotes: 0

jaromrax
jaromrax

Reputation: 283

As @anubhava suggests, this modification works for me:

#!/bin/bash
read currVal
cnt=1
val=$currVal
while [[ -n $val ]]
    do
    echo enter next
    read val
    if [[ $val == $currVal ]]; then
        cnt=$((cnt+1))
    else
        echo "$currVal occurs $cnt times"
    exit
    fi
done

Upvotes: 1

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