如何将jQuery变量传递给PHP变量?

如何在不刷新页面的情况下将变量从jQuery传递到PHP?当我单击一个复选框时,我想将一个变量从jQuery传递到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" );

// bValid = bValid && checkLength( name, "username", 3, 16 );

// bValid = bValid && checkRegexp( name, /^[a-z]([0-9a-z_])+$/i, "Username may consist of a-z, 0-9, underscores, begin with a letter." );

if ( bValid ) {

processDetails();

return false;

}

},

Cancel: function() {

$( this ).dialog( "close" );

$('input[name=opendialog]').attr('checked', false);

}

}

});

$('.opendialog').click(function() {

$('.mydialog').dialog('open');

closedialog = 0;

});

回答:

Ajax可以做到这一点。谷歌它,并检查api.jquery.com并查看ajax函数,.ajax(),.post(),.get(),.load()等。

关于您的特定问题,这是您将要做的:

//Javascript file

$("input[type=checkbox]").click(function () {

$.post('my_ajax_receiver.php', 'val=' + $(this).val(), function (response) {

alert(response);

});

});

//PHP file my_ajax_receiver.php

<?php

$value = $_POST['val'];

echo "I got your value! $value";

?>

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