网络知识之IPV6（二）

最近在处理IPV6的校验，有一个需求：要求能校验出同一个IPV6地址的简写地址和完整地址，这就需要我们把简写的IPv6地址还原，同时还能把ipv6地址简化。

下面分享给一个我使用的JS方法：

 |   // 简写ipv6
    ipv6AbbreFunc () {
      // 简写ipv6,只针对0操作：
      // 1.4位16进制数的段中高位0可以省略，
      // 2.相邻全0段，简写成::，双冒号只有一次
      let value = '1000:0000:0001:0000:0001::0000'
      // 补全ipv6地址
      let fullIPv6 = this.ipv6FillFunc(value)// abcd:0000:0000:0000:0001:0000:0000::0001
      console.log(fullIPv6)
      // 4位16进制数的段中高位0可以省略，
      let v6Arr = fullIPv6.split(':')
      v6Arr.forEach((item, index) => {
        v6Arr[index] = v6Arr[index].replace(/^0{1,3}/, '')
      })// v6Arr ['abcd','0','0','0','1','0','0','1']
      console.log(v6Arr)
      // 查找要简写的全0段的位置
      let v6Arr2 = v6Arr
      v6Arr2.forEach((item, index) => {
        if (v6Arr2[index] != '0') {
          v6Arr2[index] = '-'
        }
      })// v6Arr2 ['-','0','0','0','-','0','0','-']
      console.log(v6Arr2)
      // 找出最长重复字符的字符
      let v6Str = v6Arr2.toString()// '-,0,0,0,-,0,0,-'
      v6Str = v6Str.replace(/,/g, '')// '-000-00-'
      console.log(v6Str)
      var strmore = this.maxRepeactStr(v6Str)
      console.log(strmore)
      let maxRepeactString = ''
      if (strmore.length === 0) {
        return value
      } else if (strmore.length >= 1) { // 有重复的0
        maxRepeactString = strmore[0]
        console.log('maxRepeactString:' + maxRepeactString)
        let location = v6Str.indexOf(maxRepeactString)
        console.log('location:' + location)
        // 替换整理
        let arr = []
        v6Arr = fullIPv6.split(':')
        v6Arr.forEach((item, index) => {
          arr[location] = '::'
          if (index >= location && index < location + maxRepeactString.length) {
            arr[index] = ''
          } else {
            arr[index] = item
          }
        })
        console.log(arr)
        let abbre = ''
        arr.forEach((item, index) => {
          if (item === '::' || item === '' || index === 0 || index === parseInt(location)) {
            abbre = abbre + '' + item
          } else {
            abbre = abbre + ':' + item
          }
        })
        abbre = abbre.replace(/::+/g, '::')
        console.log(abbre)
        return abbre
      }
    },
    // 补全ipv6
    ipv6FillFunc (value) {
      let perlipv6regex = /^\s*((([0-9A-Fa-f]{1,4}:){7}([0-9A-Fa-f]{1,4}|:))|(([0-9A-Fa-f]{1,4}:){6}(:[0-9A-Fa-f]{1,4}|:))|(([0-9A-Fa-f]{1,4}:){5}(((:[0-9A-Fa-f]{1,4}){1,2})|:))|(([0-9A-Fa-f]{1,4}:){4}(((:[0-9A-Fa-f]{1,4}){1,3})|:))|(([0-9A-Fa-f]{1,4}:){3}(((:[0-9A-Fa-f]{1,4}){1,4})|:))|(([0-9A-Fa-f]{1,4}:){2}(((:[0-9A-Fa-f]{1,4}){1,5})|:))|(([0-9A-Fa-f]{1,4}:){1}(((:[0-9A-Fa-f]{1,4}){1,6})|:))|(:(((:[0-9A-Fa-f]{1,4}){1,7})|:)))(%.+)?\s*$/
      if (value !== '' && !perlipv6regex.test(value)) {
        return false
      }
      let v6Arr = value.split(':')
      if (v6Arr[0] === '') {
        v6Arr.shift()
      } else if (v6Arr[v6Arr.length - 1] === '') {
        v6Arr.pop()
      }
      // 转换点分十进制的v4写法
      let map10To16 = {
        0: '0',
        1: '1',
        2: '2',
        3: '3',
        4: '4',
        5: '5',
        6: '6',
        7: '7',
        8: '8',
        9: '9',
        10: 'a',
        11: 'b',
        12: 'c',
        13: 'd',
        14: 'e',
        15: 'f'
      }
      if (v6Arr.length > 1 && v6Arr[v6Arr.length - 1].indexOf('.') !== -1) {
        let ten = v6Arr.pop()
        let tenArr = ten.split('.')
        tenArr.forEach(tenItem => {
          v6Arr.push('00' + map10To16[Math.floor(tenItem / 16)] + map10To16[tenItem % 16])
        })
      }
      let fillStr = ''
      if (v6Arr.length !== 8) {
        let length = 9 - v6Arr.length
        for (let i = 0; i < length; i++) {
          fillStr += '0000:'
        }
      }
      let num16 = ''
      v6Arr.forEach((e) => {
        e = e.toLocaleLowerCase()
        if (e !== '') {
          num16 += (e.length === 4 ? e : '0'.repeat(4 - e.length) + e) + ':'
        } else {
          num16 += fillStr
        }
      })
      return num16.substr(0, 39)
    },
    // 找到最长的重复子串
    maxRepeactStr (str) {
      let reg = /(\w)\1+/g
      let a = []
      let max = ['']
      while ((a = reg.exec(str)) != null) {
        if (max[0].length < a[0].length) {
          max[0] = a[0]
        } else if (max[0].length == a[0].length) {
          max.push(a[0])
        }
      }
      return max
    } |