如何将JQuery变量发布到PHP？

我有如下选择字段：

        <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>

我有如下jquery可以在选择新值时获取所选值：

$("#year").change(function() {
    input_year =  parseInt($(this).val());
});

我的问题是，每次值更改时，如何将'input_year'变量发布到PHP？

最终目标是在PHP中使用$_POST['year']进行操作，并将结果返回给JavaScript，例如echo "var res = " . $_POST['year'] . ";";。

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

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 */
    });
});

$("#year").change(function() {
    var input_year =  parseInt($(this).val());

    $.post('server.php',{year:input_year}, function(result,status){});
});

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);
   }
});