如何将json对象值传递给php dinamically？

Question

Hi I'm newbie in JSON and Ajax and my question is probably quite stupid but when learning, also the stupid questions are fundamental.

I need to pass two parameters via Ajax (giorno and periodo),

for example 'giorno' = 2017-05-10 and 'periodo' = 2:

$.ajax({
    type:'POST',
    data: JSON.stringify({
    giorno: $('#dataselezionata').val(),
    periodo: $('input:radio[name=periodo]:checked').val()
    }),
    contentType: "application/json; charset=utf-8",
    dataType : 'json',
    url:'http://www.rinnovipatenti.com/index2.php?a=prenotazione' 
    });

The JSON object passed perfectly and the result in Firebug Console is:

{"giorno":"2017-05-10","periodo":"2"}

If I try to manually copy and paste the object on the php page like this:

<?
$json       = '{"giorno":"2017-05-10","periodo":"2"}'; //pasted manually

$json       = json_decode($json, TRUE);

$giorno = $json['giorno'];
$periodo    = $json['periodo'];

echo"$giorno"; //2017-05-10
echo"$periodo"; //2
?>

the two echoes show me the values. OK!

My problem start and stop here . I'm not able to bring the JSON object to be decoded.

I'm quite sure is a stupid solution but I don't know how to do that.

I need to create a function that wrap the Ajax call and then call the function in json_decode??

PS I also tried to simply get the values with "$_POST['giorno']" etc.. instead of using JSON but without success.

Can someone help me please? Thank you for your patience.

UPDATE 10/05/2017

Hi I've followed your suggestions so I tried one time more to simplify the code like this:

$('input:radio[name=periodo]').change(function() {

var giorno = document.getElementById("dataselezionata").value; // from datepicker
var periodo = $('input:radio[name=periodo]:checked').val(); // from radio button

var post_data = ("giorno="+giorno+"&periodo="+periodo);

$.ajax({
    type:'GET',
    data: post_data,
    url:"http://www.rinnovipatenti.com/prenota/prenotazione.php",
    });

if (this.value == '1') {
            $('#orario').show();
            $('#orari-mattina').show();
            $('#orari-pomeriggio').hide();
            }
else if (this.value == '2') {
            $('#orario').show();
            $('#orari-pomeriggio').show();
            $('#orari-mattina').hide();
        }

using GET method instead of the POST one and in the PHP page prenotazione.php the code now is:

<?
$giorno = $_GET['giorno'];
$periodo    = $_GET['periodo'];   

echo"$giorno";
echo"$periodo";
?>

In Firebug console the parameters are ok

giorno 2017-05-10
periodo 2

the formatted link is:

http://www.rinnovipatenti.com/prenota/prenotazione.php?giorno=2017-05-10&periodo=2

the html console preview works correctly but the page don't. I'm in the corner! Is very strange. I have only one doubt: can I send data by GET/POST method to the same page where the data are collected? In other word can the page foo.php send data to foo.php like a cycle without refresh? And the ajax call could be wrapped inside the .change function or must be outside?

Answer 1

$.post( "http://www.rinnovipatenti.com/index2.php?a=prenotazione", {
    giorno: $('#dataselezionata').val(),
    periodo: $('input:radio[name=periodo]:checked').val()
    } );

you do not need to stringify your JSON

on PHP side you just use

$giorno = $_POST['giorno'];
$periodo    = $_POST['periodo'];

to get the values

you can use the following function. you have it already. it worked fine for me.

$('input:radio[name=periodo]').change(function() {

    var giorno = '2017-05-15';
    var periodo = '2';

    $.post( 'http://www.rinnovipatenti.com/index2.php?a=prenotazione', {
        giorno: giorno,
        periodo: periodo
        });
    /*....*/
}); 

Answer 2

Method 2

You don't have to stringify the JSON object

$.ajax({
    type:'POST',
    data: {
    giorno: $('#dataselezionata').val(),
    periodo: $('input:radio[name=periodo]:checked').val()
    },
    contentType: "application/json; charset=utf-8",
    dataType : 'json',
    url:'http://www.rinnovipatenti.com/index2.php?a=prenotazione' 
    });

So you php code will be like this

<?


$giorno = $_POST['giorno'];
$periodo    = $_POST['periodo'];

echo"$giorno"; //2017-05-10
echo"$periodo"; //2
?>

Method 2

If you want to stringify the JSON object then put it in a key

$.ajax({
    type:'POST',
    data: 'data='+JSON.stringify({
    giorno: $('#dataselezionata').val(),
    periodo: $('input:radio[name=periodo]:checked').val()
    }),
    contentType: "application/json; charset=utf-8",
    dataType : 'json',
    url:'http://www.rinnovipatenti.com/index2.php?a=prenotazione' 
    });

So you php code will be like this

<?
$json       = $_POST['data'];

$json       = json_decode($json, TRUE);

$giorno = $json['giorno'];
$periodo    = $json['periodo'];

echo"$giorno"; //2017-05-10
echo"$periodo"; //2
?>

Answer 3

When you send application/json payload to php to access that payload use:

$json = json_decode(file_get_contents('php://input'), TRUE);

As mentioned in other comments and answers if you stay with $.ajax default form encoding and don't json stringify the data then use $_POST