John Davidson

Message:

I have this code in an old, unmaintened woocommerce plugin, for which the author is not around any more and it throws the following error:

PHP Parse error: syntax error, unexpected '<' in /var/www/html/wp-content/plugins/woobrdg/api.php(24) : eval()'d code on line 1"

Any suggestions on how to rectify it?

public function custom_json_decode( $json ){



    $comment = false;

    $out = '$x=';

    for ($i=0; $i<strlen($json); $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 ( object )$x;



}

---

My Answer:
The issue with this eval usage is that it is vulnerable to code injection attacks. The code is directly evaluating the input JSON string without proper validation or sanitization, which can allow an attacker to execute arbitrary code.

To rectify this issue, you should consider using a safer method to decode the JSON string, such as the built-in json_decode function in PHP. This function will parse the JSON string and return the corresponding PHP data structure without executing any potentially harmful code.

Here is an updated version of the custom_json_decode function using json_decode:

php
public function custom_json_decode( $json ){
$decoded = json_decode($json);
if ($decoded === null) {
throw new Exception('Invalid JSON string');
}
return $decoded;
}


By using json_decode, you can safely parse the JSON string without the risk of code injection.