AWK，count变量不返回0

Question

所以我有一个在Unix/Linux中使用AWK的程序即将完成！我必须返回计数、总和、最大值、最小值和平均值。当找到数字时，程序就会工作。但是，如果找不到数字并且count不迭代...我不能返回0。我得到类似于"1a2,5“之类的结果。这是我的代码，

#! /bin/awk -f

{
sum += $1
}

/[0-9]+/{ 


if (NR >= 1){
    if ($0 != 0 &&  NF == 1){
        if ($0 !~ /[A-Za-z]/){

             min = (NR==1 || $1<min ? $1 : min)
             max = (NR==1 || $1>max ? $1 : max)
             mean = sum/NR

        count+=1  
        }
    } else if ($0 == 0)
    {exit(0)
}
}
  }


END{
printf("# items: %d \n",count)
printf("Total:   %lf\n", sum)
printf("Maximum: %lf\n", max)
printf("Minimum: %lf\n", min)
printf("Average: %lf\n", mean)
}

我忘记了什么/需要更改，这样如果什么都没有找到，count就可以返回0。谢谢

Answer 1

#! /bin/awk -f

# Nothing to do
NF > 1 {
    next
}

# now $0 == $1

# Only with numbers
#$1 ~ /^[0-9]+$/ {
# or
$1 ~ /^[0-9]+(\.[0-9]+)?$/ {
    # exit if zero found
    if ($1 == 0) {
        exit 0
    }
    # compute
    sum   += $1
    min    = (NR==1 || $1<min ? $1 : min)
    max    = (NR==1 || $1>max ? $1 : max)
    mean   = sum/NR
    count += 1
}

END {
    printf("# items: %d \n", count)
    printf("Total:   %lf\n", sum)
    printf("Maximum: %lf\n", max)
    printf("Minimum: %lf\n", min)
    printf("Average: %lf\n", mean)
}