Using json_encode() and json_decode() in PHP4

I use json_encode() a lot for AJAX calls. Teamed with jQuery’s $.getJSON(), it’s too convenient not to use. Unfortunately, json_encode() doesn’t come standard until PHP 5.2. To add insult to injury, many current *nix distros don’t include PHP 5.2 in their official repositories yet.

So, if you’re using PHP4 download JSON’s json_encode and json_decode for PHP4 (which automatically degrade for PHP5, thus not breaking your app during an upgrade) here:

json_encode() example 1


$a = json_encode( array( 'a'=>1, '2'=>2, 'c'=>'I <3 JSON' ) );

echo $a;
// Outputs: {"a":1,"b":2,"c":"I <3 JSON"}

$b = json_decode( $a );

echo "$b->a, $b->b, $b->c";
// Outputs: 1, 2, I <3 JSON

json_encode() / json_decode() example 2

In Javascript, consuming input in JSON format is as easy as:

eval("var decoded_data = " + encoded_data);

With JSON-PHP, it can be almost as easy on the server-side, too:

// create a new instance of Services_JSON
$json = new Services_JSON();

// convert a complex value to JSON notation
$value = array(1, 2, 'foo');
$output = $json->encode($value);

// accept incoming POST data
$value = $json->decode($input);

Using json_encode() and json_decode() in PHP4 with Arrays

You can use json_encode() with arrays or multi-dimensional arrays. When you are ready to output it as json, just call

echo json_encode($yourArray);

You can also use this method with jQuery’s $.ajax call, and specify the data type as JSON, although jQuery will automatically try and detect the data type, so that may not even be necessary.

source: mike.teczno

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10 Responses to Using json_encode() and json_decode() in PHP4

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