jQuery - Serialize Form Post Values - php

Question

How do I want to post a form using jquery serialize function? I tried to post the form value but on the php part, the value is not shown. Below are my codes:
html
<form name="myform">
ID : <input type="text" name="id_staff" id="id_staff">
<select name="sort" id="sort">
<option value="0">Choose Status</option>
<option value="1">All</option>
<option value="2">Pending</option>
<option value="3">Approve</option>
<option value="4">Not Approve</option>
</select> <input type="button" id="submit" value="Papar" />
<div id="loader"></div>
</form>
jQuery
$(document).on("click", "#submit", function(e){
e.preventDefault();
var sort = $("#sort").val(),
id_staff = $("#id_staff").val(),
data = $('form').serialize();
$.post('result.php',
{
data : data
}, function(data){
$("#loader").fadeOut(400);
$("#result").html(data);
});
});
PHP
if(isset($_REQUEST["sort"])){
$sort = $_REQUEST['sort'];
$id_staff = $_REQUEST['id_staff'];
echo "Your Id : $id_staff <p/>";
echo "You choose : $sort";
}
If I console.log(data), I get: id_staff=12345&sort=1

Answer 1

Your server is receiving a string that looks something like this (which it should if you're using jQuery serialize()):
"param1=someVal&param2=someOtherVal"
...something like this is probably all you need:
$params = array();
parse_str($_GET, $params);
$params should then be an array that contains all the form element as indexes

Answer 2

If you are using .serialize, you can get rid of this:
var sort = $("#sort").val(),
id_staff = $("#id_staff").val(),
You data will be available as follows with .serialize:
your-url.com/sort=yoursortvalue&id_staff=youridstaff
It should be:
$(document).ready(function(e) {
$("#myform").submit(function() {
var datastring = $( this ).serialize();
$.post('result.php',
{
data : datastring
}, function(data){
$("#loader").fadeOut(400);
$("#result").html(data);
});
})
})
On PHP side you simple need to access it using the $_GET['sort'].
Edit:
To view the data, you should define a div with id result so that the result returned is displayed within this div.
Example:
<div id="result"></div>
<form name="myform">
ID : <input type="text" name="id_staff" id="id_staff">
<select name="sort" id="sort">
<option value="0">Choose Status</option>
<option value="1">All</option>
<option value="2">Pending</option>
<option value="3">Approve</option>
<option value="4">Not Approve</option>
</select> <input type="button" id="submit" value="Papar" />
<div id="loader"></div>
</form>

Answer 3

I am able to do it this way:
jQuery
<script type="text/javascript">
$(document).ready(function() {
var form = $("#myform");
$("#myform").submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: 'result.php',
data: form.serialize(),
success: function(response) {
console.log(response);
$("#result").html(response);
},
error: function() {
alert('Error Submitting');
}
})
})
})
</script>
PHP
if(isset($_POST["id_staff"])){
$sort = $_POST['sort'];
$id_staff = $_POST['id_staff'];
echo "<p>Your Id : $id_staff</p>";
echo "You choose : $sort";
}
Do give a comment if it need improvement or better solution.