JavaScript IP 范围验证第二个 ip 大于第一个 ip

Question

您好，我是一名 javascript 菜鸟，正在为 IP 范围编写验证器。例如，1.1.1.1-2.2.2.2 是一个有效范围，但我想确保第一个 IP 不大于第二个 IP。 2.2.2.2-1.1.1.1 应返回 false。解决这个问题的最佳策略是什么，我应该尝试将每个 IP 字符串转换为整数映射，然后将它们一一比较吗？我应该迭代 for 循环 1-4 并比较 .然后是第二个等等。

ipRangeValidator: function(value)
{
    if (!value)
        return true;
    var split = value.split('-');
    if (!split || split.length > 2 || split.length === 0) {
        return 'Input only an IP or single IP Range';
    }

    var ips = [];
    for (var x = 0; x < split.length; ++x) {
        var ip = split[x];
        if (isValidIpV6Address(ip)) { 
        } else if (isValidIpV4Address(ip)) {
        } else {
            return ip + ' is not a valid IP address';
        }
        ips.push(ip);
    }
    if (ips.length > 1) {
       if(true) {            //The Line I am struggling with
           return 'The first IP is greater than the second IP';
       }
    }
    return true;
}

Answer 1

像这样的东西对你有用吗？

const validRange = '1.1.1.1-2.2.2.2',
      invalidRange = '2.2.2.2-1.1.1.1',
      anotherInvalidRange = '1.1.3.1-1.1.2.1',
      invalidIp = '1.1.300.1-1.1.1.1',
      
      isRangeValid = range => {
        const [rangeStart, rangeEnd] = range.split('-'),
              [s1,s2,s3,s4] = rangeStart.split('.'),
              [e1,e2,e3,e4] = rangeEnd.split('.')
        return [s1,s2,s3,s4].some(o => o>255 || o<0) || [e1,e2,e3,e4].some(o => o>255 || o<0) ?
          'range contains invalid IP' :
          s1>e1 || s2>e2 || s3>e3 || s4>e4 ?
          'range start is greater than range end' :
          'range is valid'
      }
      
console.log(isRangeValid(validRange))
console.log(isRangeValid(invalidRange))
console.log(isRangeValid(anotherInvalidRange))
console.log(isRangeValid(invalidIp))

.as-console-wrapper{min-height:100%;}