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如何在没有页面刷新的情况下将变量从jQuery传递给PHP？当我点击一个复选框时,我想将一个变量从jQuery传递给PHP.我也在使用formdialog.

我的PHP代码

<?php
echo "<input name='opendialog' type='checkbox' class='opendialog' onclick='countChecked()' value=".$taskid." ?>" /> </td>"
?>

我的javascript代码

function countChecked() {
  var n = $("input:checked").length;

  var allVals = [];
   $('input:checkbox:checked').each(function() {
   allVals.push($(this).val());

   });
   $('.sel').text(allVals+' ');
   $('.select1').val(allVals);
   alert(allVals);

    <?php $taskidj=$rowtask['taskID'];
   // echo "aaa...".$rowtask['taskID']; ?>     

}

$(":checkbox").click(countChecked);


// my jquery code


        $('.mydialog').dialog({

            bgiframe: true,
            autoOpen: false,
            modal: true,
            width: 700,
            height:500,
            resizable: false,
            open: function(){closedialog = 1;$(document).bind('click', overlayclickclose);},
            focus: function(){closedialog = 0;},
            close: function(){$(document).unbind('click');},
            buttons: {
                Submit: function(){
                var bValid = true;
            //  allFields.removeClass( "ui-state-error" …