PHP 如何实现ip2cidr（生成多个cidr）-php教程-PHP中文網

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PHP 如何实现ip2cidr（生成多个cidr）

WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWB

Jun 23, 2016 pm 01:52 PM

php 多個產生

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

回复讨论(解决方案)

function ip2cidr($ip_start,$ip_end) {
if(long2ip(ip2long($ip_start))!=$ip_start or long2ip(ip2long($ip_end))!=$ip_end) return NULL;
$ipl_start=(int)ip2long($ip_start);
$ipl_end=(int)ip2long($ip_end);
if($ipl_start>0 && $ipl_end else $delta=$ipl_end-$ipl_start;
$netmask=str_pad(decbin($delta),32,"0","STR_PAD_LEFT");
if(ip2long($ip_start)==0 && substr_count($netmask,"1")==32) return "0.0.0.0/0";
if($delta0 && $delta%2==0)) return NULL;
for($mask=0;$mask if(substr_count($netmask,"0")!=$mask) return NULL;
return "$ip_start/$mask";
} 这个函数不能生成多个cidr

echo ip2cidr('1.120.0.0', '1.127.255.255'); //1.120.0.0/13function ip2cidr($ip_start,$ip_end) {  if(long2ip(ip2long($ip_start))!=$ip_start or long2ip(ip2long($ip_end))!=$ip_end) return NULL;   $ipl_start = ip2long($ip_start);  $ipl_end = ip2long($ip_end);  if($ipl_start>0 && $ipl_end<0) $delta = ($ipl_end + 4294967296) - $ipl_start;  else $delta = $ipl_end - $ipl_start;  $netmask = str_pad(decbin($delta), 32, "0", STR_PAD_LEFT);  if(ip2long($ip_start)==0 && substr_count($netmask,"1")==32) return "0.0.0.0/0";  if($delta<0 or ($delta>0 && $delta%2==0)) return NULL;  for($mask=0; $mask<32; $mask++) if($netmask[$mask]==1) break;  if(substr_count($netmask,"0")!=$mask) return NULL;  return "$ip_start/$mask";}

登入後複製

echo ip2cidr('1.120.0.0', '1.127.255.255'); //1.120.0.0/13function ip2cidr($ip_start,$ip_end) {  if(long2ip(ip2long($ip_start))!=$ip_start or long2ip(ip2long($ip_end))!=$ip_end) return NULL;   $ipl_start = ip2long($ip_start);  $ipl_end = ip2long($ip_end);  if($ipl_start>0 && $ipl_end<0) $delta = ($ipl_end + 4294967296) - $ipl_start;  else $delta = $ipl_end - $ipl_start;  $netmask = str_pad(decbin($delta), 32, "0", STR_PAD_LEFT);  if(ip2long($ip_start)==0 && substr_count($netmask,"1")==32) return "0.0.0.0/0";  if($delta<0 or ($delta>0 && $delta%2==0)) return NULL;  for($mask=0; $mask<32; $mask++) if($netmask[$mask]==1) break;  if(substr_count($netmask,"0")!=$mask) return NULL;  return "$ip_start/$mask";}

登入後複製

版主大人您写的和我2楼一样解决不了1.120.0.0 1.159.255.255 这个IP区间生成CIDR的问题

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

而
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

收到谢谢版主大人我好好琢磨下

ip2cidr("1.40.0.0","1.44.255.255");
00000000000001001111111111111111
echo long2ip(bindec('111111111111111111111')+ip2long('1.40.0.0')); //1.71.255.255 超出了指定IP的范围

0000000000000100 1111111111111111
要这样取

拆分理应从小区域开始
上来就是一大块，肯定是不合适的

谢谢您我摸索下

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

结束

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

登入後複製

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

谢谢版主大人尤其认识到自身不足还要好好努力啊

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