Q:

How to handle JSON object in php

I figure it out to create a JSON object in jquery 
function form2JSON(form){
    var info_ser = $('#'+form).serialize();
    var data = info_ser.split('&');
    var output = {};
    $.each( data, function( i, l ){
        var data_input = l.split('=');
        output[data_input[0]] = data_input[1];
     });
    return output;
}

But now in PHP I dont know how to handle the JSON.... ? 

A:

$.ajax({
    type: "POST",
    url: "./modelo/.php",
    data: {
        act: 'mc',
        ent: 'modelos',
        json: jsonobject
    },
    dataType: 'json',
    cache: false
}).done(function(data) {});

Then you can access your data in php by:
json_decode($_POST['json']);

You actually forgot to add a key in data for your jsonobject.

