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