How can I decode json in PHP 5.1?

问题:

json_decode function is not part of PHP 5.1, so I cannot use it. Is there any other function for this version?

回答1:

Before PHP 5.2, you can use the JSON PECL extension.

In fact, it is the extension that has been integrated into PHP 5.2 (quoting) :

As of PHP 5.2.0, the JSON extension is bundled and compiled into PHP by default.

Some other solutions would be to use some component developped in PHP.

Some time ago (about one year ago), I used the Zend_Json component of Zend Framework, with PHP 5.1.

Even if Zend Framework requires PHP 5.2, that component can be extracted (I think it depends only on one other component -- maybe Zend_Exception, or something like that) -- and one year ago, it was possible to use it with PHP 5.1.

The official JSON website also links to several PHP-based or PHP-oriented components -- you might want to take a look at that list.

回答2:

I ran into the same issue running PHP 5.1.6, but I couldn't upgrade or install extensions on my client's server. To make matters worse, the JSON.org site was down when I needed a solution but fortunately this file on Google Code worked perfectly! I would have preferred to actually define json_encode/json_decode, but calling fromJSON() worked just fine.

http://code.google.com/p/simplejson-php/

回答3:

You're seeing this error because you have a php version earlier than 5.2.0. These functions are included by default in php 5.2.0 and later.

PHP Fatal error:  Call to undefined function json_encode()

You can install the PECL extension by running:

pecl install json

It will compile, then add this to your php.ini file: (mine is in /etc/php5/apache2)

extension=json.so

Then restart apache.

回答4:

In my server i can't install JSON PECL extension, because it causes a problem with zend_json that is used in another app. So i found this script that works perfectly.

jsonwrapper: json_encode for earlier versions of PHP 5.x

PHP 5.2 adds the json_encode function, which turns almost any PHP data structure into valid JavaScript code. Hashes, arrays, arrays of hashes, whatever.

Unfortunately a lot of Linux distributions are still shipping with PHP 5.1.x.

jsonwrapper implements the json_encode function if it is missing, and leaves it alone if it is already present. So it is nicely future-compatible.

Just add:

require 'jsonwrapper.php';

http://www.boutell.com/scripts/jsonwrapper.html

回答5:

The Zend framework has Zend_Json. At least it used to a couple of years ago.

http://framework.zend.com/download

You can just pull out the JSON library and use it in a standalone manner.

回答6:

code

<?php 
if ( !function_exists('json_decode') ){ 
function json_decode($json) 
{  
    // Author: walidator.info 2009 
    $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] == '"')    $comment = !$comment; 
    } 
    eval($out . ';'); 
    return $x; 
}  
} 
?>

warning

this is untested, I found it on the internet

link

http://www.php.net/manual/en/function.json-decode.php#91216

回答7:

I ran into problems with the Services_Json extension on PHP 5.1.3, so I found the following library:

https://github.com/alexmuz/php-json

It is under LGPL, and after a very quick look does not seem to eval input.

回答8:

You can use jsonwrapper library...

jsonwrapper implements the json_encode function if it is missing, and leaves it alone if it is already present. So it is nicely future-compatible.

Download here: jsonwrapper

To use just do:

require ("jsonwrapper.php");


$data = array('idx1' => 'foo', 'idx2' => 'bar');

echo json_encode($data);

echo json_decode($data);