Json在PHP中解码

Question

I am trying to decode JSON in php I sended using an ajax call in javascript (jquery).

The javascript code:

    var editSchedule_data = {
        'action'            : 'editSchedule',
        'eventid'           : $('#eventid').val(),
        'pixels_per_minute' :$('#pixels_per_minute').val(),
         artists: {
            artist: []
        }
    }

    $('.artist').each(function(index) {
        var id=$(this).attr('id');
        id=id.split('_');
        editSchedule_data.artists.artist.push({
            'artistid'    : id[0],
            'stageid'     : id[1],
            'left'        : $(this).offset().left-$('.artists').offset().left,
            'width'       : $(this).width()
        });
    });

    $.ajax({
        type :   "POST",
        url  :   "callback.php",
        data :   editSchedule_data,
        dataType :   "json",
        async    :   true,
        success   :   function(json)  {
            if(json.success)  {
                showSucces(json.message);

            }
            else{
                showError(json.message);
            }
        },
        error: function(error){
            alert("An error occurred: "+error.message);
        }
    });

The php-code:

$clean = sanitize($_POST);

echo(json_encode($clean['artists']),
json_last_error());

echo(json_decode($clean['artists']),
json_last_error());

My output: encode:

{"artist":[{"artistid":"4","stageid":"3","left":"360","width":"240"},{"artistid":"3","stageid":"4","left":"120","width":"240"},{"artistid":"1","stageid":"5","left":"120","width":"180"},{"artistid":"2","stageid":"5","left":"300","width":"120"},{"artistid":"5","stageid":"6","left":"480","width":"120"}]}
0

decode:

0

Can someone tell me how to get the decode function to work?

Answer 1

Why are you trying to use json_decode there? You already have the data as array in $_POST.That's why json_decode fails and retuns NULL...it expects a properly formated string and you are passing it an array. If the dataType : "json" parameter confuses you, it specifies the type of the data you are expecting back from the server not the type of the data you are sending. You shoud simply process the data from $_POST, create your response , apply json_encode on it and echo the resulted string.

Answer 2

json_decode($clean['artists']); is giving you an object, that's why echoing it doesn't show anything. Try print_r() to see what's inside the object:

// Will show you the whole PHP object
print_r( json_decode( $clean['artists'] ) ); 

Answer 3

it is much better if you will use this. when fetching data from the database

$.getJSON("callback.php",function(json){
$('#div to update').html(json[0].databasefield); // 0 if query is only 1 result
});                                 // when there are many of them use for loop 

on your php you should encode it using json_encode(array values);