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我想把2 datePicky jQuery UI。错误:当我加载页面时,datePickers没有显示。但是,如果我发布表单,它会显示它们。datePicker jQuery UI - 没有显示正常的负载,但在窗体上加载
这里是我的代码:
<?php
//...
if ($_POST){
$dateFrom=getInput('dateFrom').' 00:00:00';
$dateTo=getInput('dateTo').' 23:59:59';
$defDate=explode('-',getInput('dateFrom'));
$defDate=$defDate[1].'/'.$defDate[2].'/'.$defDate[0];
$employeeID=getInput('employeeID');
}else{
$dateFrom=date('Y-m-d',strtotime('-1 day'));
$dateTo=date('Y-m-d').' 23:59:59';
$defDate=date('m/d/Y',strtotime('-1 day'));
$employeeID=0;
}
//================================== calendar widget ==================================
echo '<script type="text/javascript">$(function() {';
echo ' $("#datepicker1").datepicker("option", "dateFormat", "yy-mm-dd");
$("#datepicker1").datepicker({
showWeek: true,
firstDay: 0,
changeMonth: true,
defaultDate: "'.$defDate.'",
changeYear: true,';
echo " altField: '#datepickera',altFormat: 'yy-mm-dd'});";
echo ' $("#datepicker2").datepicker("option", "dateFormat", "yy-mm-dd");
$("#datepicker2").datepicker({
showWeek: true,
firstDay: 0,
changeMonth: true,
changeYear: true,';
echo " altField: '#datepickerb',altFormat: 'yy-mm-dd'});";
echo '});';
echo '</script>';
echo '<form action="" method="post">';
echo '<table>';
echo '<tr><td width="250">From Date<input type="text" id="datepickera" name="dateFrom" value="" /><br /><div id="datepicker1"></div></td>';
echo '<td width="250">To Date<input type="text" id="datepickerb" name="dateTo" value="" /><br /><div id="datepicker2"></div></td>';
echo '<td>'. __('Choose Employee','j-v3').':<br /><select name="employeeID" /><option value="0">--- All Employees ---</option>';
echo '</select> <input type="submit" value="Get Data" /></td>';
echo '</tr>';
echo '</table>';
echo '</form>';
//...
?>
我使用jQuery 1.6.2 & jQuery的UI 1.8.15
请帮助。如何总是显示datePickers?
http://jqueryui.com/datepicker/ – Arshad 2013-02-26 08:05:36
@arshad我在这里问,因为问题无法通过自述解决。所以没有人知道? – 2013-02-27 01:49:39