Home php教程 php手册 php5.2以下版本无json_decode函数的解决方法

php5.2以下版本无json_decode函数的解决方法

Jun 13, 2016 am 09:34 AM
decode json

今天写代码的时候,需要用到json_decode函数,发现php5.2以前的版本没有集成这个函数,不过我们可以通过自定义函数实现。

复制代码 代码如下:


function json_decode2($json)
{
$comment = false;
$out = '$x=';

for ($i=0; $i{
if (!$comment)
{
if (($json[$i] == '{') || ($json[$i] == '[')) $out .= ' array(';
else if (($json[$i] == '}') || ($json[$i] == ']')) $out .= ')';
else if ($json[$i] == ':') $out .= '=>';
else $out .= $json[$i];
}
else $out .= $json[$i];

if ($json[$i] == '"' && $json[($i-1)]!="\\") $comment = !$comment;
}

    eval($out . ';');
return $x;
}

不过这个返回的是Array

要返回object 则要用到 service_json类了

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