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这段PHP代码如何写成能输出的结果和JS一样

Jun 13, 2016 pm 12:08 PM

amp lt nbsp return while

这段PHP代码怎么写成能输出的结果和JS一样？
下面是我自己转成PHP代码

 function na($a) {  if (!$a) return "";  $a = (string)$a;  $h=array(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,62,-1,-1,-1,63,52,53,54,55,56,57,58,59,60,61,-1,-1,-1,-1,-1,-1,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,-1,-1,-1,-1,-1,-1,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,-1,-1,-1,-1,-1);  $i = strlen($a);  $e="";  for ($f = 0;$f<$i;){  do {$aa=$a[$f++];$c = $h[ord($aa) & 255];}while ($f < $i && -1 == $c);  if (-1 == $c) break;  do {$aa2=$a[$f++];$b = $h[ord($aa2) & 255];} while ($f < $i && -1 == $b);  if (-1 == $b) break;  $e += (string)chr($c << 2 | ($b & 48) >> 4);  do{ $aa3=$a[$f++];$c=ord($aa3)&255;if(61==$c)return $e;$c=$h[$c];}while($f<$i&&-1==$c);  if (-1 == $c) break;  $e += (string)chr(($b & 15) << 4 | ($c & 60) >> 2);  do{ $aa4=$a[$f++];$b=ord($aa4)&255; if(61==$b)return $e;$b=$h[$b];}while($f<$i&&-1==$b);  if (-1 == $b) break;  $e += (string)chr(($c &3) << 6 | $b); }  //echo $e; return $e; } 

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下面是原JS代码

 function na(a) {  if (!a) return "";  var a = a.toString(), c, b, f, i, e, h = [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 62, -1, -1, -1, 63, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, -1, -1, -1, -1, -1, -1, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, -1, -1, -1, -1, -1, -1, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, -1, -1, -1, -1, -1]; i = a.length; f = 0; for (e = ""; f < i;) {  do c = h[a.charCodeAt(f++) & 255]; while (f  if (-1 == c) break;  do b = h[a.charCodeAt(f++) & 255]; while (f  if (-1 == b) break; e += String.fromCharCode(c << 2 | (b & 48) >> 4);  do { c = a.charCodeAt(f++) & 255;  if (61 == c) return e; c = h[c] } while (f  if (-1 == c) break; e += String.fromCharCode((b & 15) << 4 | (c & 60) >> 2);  do { b = a.charCodeAt(f++) & 255;  if (61 == b) return e; b = h[b] } while (f  if (-1 == b) break; e += String.fromCharCode((c & 3) << 6 | b) }  return e } 

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我用这个字符串做测试：NAXRSg4bL7zd0vjF8+JxVNXyuUBv1wnLURc=

在PHP测试结果是：4
但在JS里面测试结果是：4?J[email protected] ?Q
不知道哪里错误了

------解决思路----------------------

function na($a) {  if (!$a) return "";  $a = "$a";  $h = [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 62, -1, -1, -1, 63, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, -1, -1, -1, -1, -1, -1, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, -1, -1, -1, -1, -1, -1, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, -1, -1, -1, -1, -1]; $i = strlen($a);  $f = 0;  for ($e = ""; $f < $i;) {  do $c = $h[ord($a{$f++}) & 255]; while ($f < $i && -1 == $c);  if (-1 == $c) break;  do $b = $h[ord($a{$f++}) & 255]; while ($f < $i && -1 == $b);  if (-1 == $b) break;  $e .= chr($c << 2 ------解决思路---------------------- ($b & 48) >> 4);  do {  $c = ord($a{$f++}) & 255;  if (61 == $c) return $e;  $c = $h[$c]; } while ($f < $i && -1 == $c);  if (-1 == $c) break;  $e .= chr(($b & 15) << 4 ------解决思路---------------------- ($c & 60) >> 2);  do {  $b = ord($a{$f++}) & 255;  if (61 == $b) return $e;  $b = $h[$b]; } while ($f < $i && -1 == $b);  if (-1 == $b) break;  $e .= chr(($c & 3) << 6 ------解决思路----------------------  $b); }  return $e; } 