Question

我正在尝试编写一个输入数字的脚本，然后输入另一个值，如果该值与输入的第一个数字相同，则会添加到cnt变量，然后让您输入另一个数字进行检查。应该重复此操作，直到两个数字不匹配，然后打印值发生的次数。我遇到循环问题，输入第二个值后，它会连续打印： 5次出现

    #!/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

Answer 1

正如@anubhava建议的那样，这种修改对我有用：

#!/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

Answer 2

计算输入值的变量：

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

BASH中的计数器计算输入数字的次数？

2 个答案: