json_encode的php实现
admin发表于 2011-02-25 09:53:44
今天coding时，需要将php中的变量返回给第三方，第一个想到就是使用json格式，方便易用。于是用了json_encode，但是发现居然报语法错误，查了手册才发现，json_enode 是在5.2.0才作为标准扩展加入php的，而我们的机器上php的版本是5.1.6，原来该扩展还未加进来。由于这个应用比较小，而且本人赖得升级php的版本，而且再手工安装这个扩展也比较麻烦，所以想到了使用php实现json_encode，原理上应该不难，找到一段代码，共享之~
代码如下：
function __json_encode( $data ) {
if( is_array($data) || is_object($data) ) {
$islist = is_array($data) && ( empty($data) || array_keys($data) === range(0,count($data)-1) );
if( $islist ) {
$json = '[' . implode(',', array_map('__json_encode', $data) ) . ']';
} else {
$items = Array();
foreach( $data as $key => $value ) {
$items[] = __json_encode("$key") . ':' . __json_encode($value);
}
$json = '{' . implode(',', $items) . '}';
}
} elseif( is_string($data) ) {
# Escape non-printable or Non-ASCII characters.
# I also put the \\ character first, as suggested in comments on the 'addclashes' page.
$string = '"' . addcslashes($data, "\\\"\n\r\t/" . chr(8) . chr(12)) . '"';
$json = '';
$len = strlen($string);
# Convert UTF-8 to Hexadecimal Codepoints.
for( $i = 0; $i < $len; $i++ ) {
$char = $string[$i];
$c1 = ord($char);
# Single byte;
if( $c1 <128 ) {
$json .= ($c1 > 31) ? $char : sprintf("\\u%04x", $c1);
continue;
}
# Double byte
$c2 = ord($string[++$i]);
if ( ($c1 & 32) === 0 ) {
$json .= sprintf("\\u%04x", ($c1 - 192) * 64 + $c2 - 128);
continue;
}
# Triple
$c3 = ord($string[++$i]);
if( ($c1 & 16) === 0 ) {
$json .= sprintf("\\u%04x", (($c1 - 224) <<12) + (($c2 - 128) << 6) + ($c3 - 128));
continue;
}
# Quadruple
$c4 = ord($string[++$i]);
if( ($c1 & 8 ) === 0 ) {
$u = (($c1 & 15) << 2) + (($c2>>4) & 3) - 1;
$w1 = (54<<10) + ($u<<6) + (($c2 & 15) << 2) + (($c3>>4) & 3);
$w2 = (55<<10) + (($c3 & 15)<<6) + ($c4-128);
$json .= sprintf("\\u%04x\\u%04x", $w1, $w2);
}
}
} else {
# int, floats, bools, null
$json = strtolower(var_export( $data, true ));
}
return $json;
}
?>