PHP 怎么实现ip2cidr(生成多个cidr)

WBOY
Lepaskan: 2016-06-13 12:04:16
asal
1474 orang telah melayarinya

PHP 如何实现ip2cidr(生成多个cidr)
各位大神有没有生成多个cidr的函数  例如1.120.0.0 1.159.255.255    生成 1.120.0.0/13   1.128.0.0/11
------解决方案--------------------
虽然拆分并不是很困难,但你如何确定拆分点呢?
比如
echo ip2cidr('1.120.0.0', '1.127.255.255'); //1.120.0.0/13
echo ip2cidr('1.128.0.0', '1.159.255.255'); //1.128.0.0/11
是一种拆法
echo ip2cidr('1.120.0.0', '1.151.255.255'); //1.120.0.0/11
echo ip2cidr('1.152.0.0', '1.159.255.255'); //1.152.0.0/13
又是一种拆法


echo ip2cidr('1.120.0.0', '1.159.255.255');
失败的原因是掩码为
00000000001001111111111111111111
其实本身并没有错,只是不能 cidr 表示而已

注意到
echo long2ip(bindec('111111111111111111111')+ip2long('1.120.0.0')); //1.151.255.255
所以那个第二种拆法是可以机器实现的,而第一种似只能手工实现


------解决方案--------------------
把问题多分析一下
00000001011101110000000000000000 1.119.0.0
00000001011110000000000000000000 1.120.0.0
00000001011110010000000000000000 1.121.0.0
00000001011110100000000000000000 1.122.0.0
00000001011110110000000000000000 1.123.0.0
00000001011111000000000000000000 1.124.0.0
00000001011111010000000000000000 1.125.0.0
00000001011111100000000000000000 1.126.0.0
00000001011111110000000000000000 1.127.0.0
00000001100000000000000000000000 1.128.0.0
00000001100000010000000000000000 1.129.0.0
00000001100000100000000000000000 1.130.0.0
00000001100000110000000000000000 1.131.0.0
00000001100001000000000000000000 1.132.0.0
00000001100001010000000000000000 1.133.0.0
00000001100001100000000000000000 1.134.0.0
00000001100001110000000000000000 1.135.0.0
00000001100010000000000000000000 1.136.0.0
00000001100010010000000000000000 1.137.0.0
00000001100010100000000000000000 1.138.0.0
00000001100010110000000000000000 1.139.0.0
00000001100011000000000000000000 1.140.0.0
00000001100011010000000000000000 1.141.0.0
00000001100011100000000000000000 1.142.0.0
00000001100011110000000000000000 1.143.0.0
00000001100100000000000000000000 1.144.0.0
00000001100100010000000000000000 1.145.0.0
00000001100100100000000000000000 1.146.0.0
00000001100100110000000000000000 1.147.0.0
00000001100101000000000000000000 1.148.0.0
00000001100101010000000000000000 1.149.0.0
00000001100101100000000000000000 1.150.0.0
00000001100101110000000000000000 1.151.0.0
00000001100110000000000000000000 1.152.0.0
00000001100110010000000000000000 1.153.0.0
00000001100110100000000000000000 1.154.0.0
00000001100110110000000000000000 1.155.0.0
00000001100111000000000000000000 1.156.0.0
00000001100111010000000000000000 1.157.0.0
00000001100111100000000000000000 1.158.0.0
00000001100111110000000000000000 1.159.0.0
00000001101000000000000000000000 1.160.0.0

------解决方案--------------------
结束

echo ip2cidr('1.120.0.0', '1.159.255.255'), PHP_EOL;<br />echo ip2cidr('1.120.0.0', '1.169.255.255'), PHP_EOL;<br />echo ip2cidr('1.120.0.0', '1.179.255.255'), PHP_EOL;<br /><br />function ip2cidr($ip_start,$ip_end) {<br />  if(long2ip(ip2long($ip_start))!=$ip_start or long2ip(ip2long($ip_end))!=$ip_end) return !trigger_error('ip 不合法', E_USER_NOTICE); <br />  $ipl_start = ip2long($ip_start);<br />  $ipl_end = ip2long($ip_end);<br />  if($ipl_start>0 && $ipl_end<0) $delta = ($ipl_end + 4294967296) - $ipl_start;<br />  else $delta = $ipl_end - $ipl_start;<br />  $netmask = sprintf('%032b', $delta);<br />  if(ip2long($ip_start)==0 && substr_count($netmask,"1")==32) return "0.0.0.0/0";<br />  if($delta<0 or ($delta>0 && $delta%2==0)) return !trigger_error("区间数量不合法 $delta", E_USER_NOTICE);<br />  for($mask=0; $mask<32; $mask++) if($netmask[$mask]==1) break;<br />  if(substr_count($netmask,"0")!=$mask) {<br />    $w = strrpos($netmask, '0') + 1;<br />    $m = pow(2, 32-$w) - 1;<br />    $ip_start = long2ip(($ipl_start & ~$m)+$m+1);<br />    return long2ip($ipl_start & ~$m) . "/$w," . ip2cidr($ip_start,$ip_end);<br />  };<br />  return "$ip_start/$mask";<br />} <br />
Salin selepas log masuk
1.120.0.0/13,1.128.0.0/11
1.120.0.0/15,1.112.0.0/12,1.128.0.0/15,1.128.0.0/13,1.136.0.0/15,1.138.0.0/11
1.120.0.0/14,1.120.0.0/13,1.128.0.0/14,1.128.0.0/12,1.144.0.0/14,1.148.0.0/11

Label berkaitan:
sumber:php.cn
Kenyataan Laman Web ini
Kandungan artikel ini disumbangkan secara sukarela oleh netizen, dan hak cipta adalah milik pengarang asal. Laman web ini tidak memikul tanggungjawab undang-undang yang sepadan. Jika anda menemui sebarang kandungan yang disyaki plagiarisme atau pelanggaran, sila hubungi admin@php.cn
Tutorial Popular
Lagi>
Muat turun terkini
Lagi>
kesan web
Kod sumber laman web
Bahan laman web
Templat hujung hadapan
Tentang kita Penafian Sitemap
Laman web PHP Cina:Latihan PHP dalam talian kebajikan awam,Bantu pelajar PHP berkembang dengan cepat!