Post JQuery variable to PHP - php

Question

So I have this select field:
<select name="year" id = "year" class="dropdown-select">
<option value="2005">All Years</option>
<option value="2013">2013</option>
<option value="2012">2012</option>
<option value="2011">2011</option>
<option value="2010">2010</option>
<option value="2009">2009</option>
<option value="2008">2008</option>
<option value="2007">2007</option>
<option value="2006">2006</option>
</select>
And I have this jquery to get the selected value any time a new value is selected:
$("#year").change(function() {
input_year = parseInt($(this).val());
});
My question is how would I go about posting the 'input_year' variable to PHP every time the value changes?
The end goal is to do something with $_POST['year'] in PHP and return the result to JavaScript e.g echo "var res = " . $_POST['year'] . ";";

Answer 1

A very simple implementation
$.ajax({
type:'POST',
url:'your_url.php',
data:'input_year='+input_year,
success: function(msg) {
console.log(msg);
}
});
data string may need a ? before it, I can't remember off the top of my head

Answer 2

Use AJAX, specifically the jQuery .post() function:
$("#year").change(function() {
input_year = parseInt($(this).val());
$.post('/path/to/file', year: input_year, function(data, textStatus, xhr) {
/*optional stuff to do after success */
});
});

Answer 3

$("#year").change(function() {
var input_year = parseInt($(this).val());
$.post('server.php',{year:input_year}, function(result,status){});
});

Answer 4

Using ajax. With jquery, it would be like this:
$.ajax({
type: 'POST',
dataType: 'text',
url: "(your php filename)",
data: $.param({'year': input_year}),
success:function(data){
//do whatever you need to do with the received result
}
error: function (request, status, error) {
alert(request.responseText);
}
});