我有一個帶有2個datepicker的html頁面,我必須在日/月/年框中選擇日期。查找直接父級jQuery
我的代碼工作正常時,我有1個日期選擇器,但在第一個總是寫日期爲2
<div class="pickergroup">
<input name="dateDay01" id="dateDay01">/
<input name="dateMonth01" id="dateMonth01">/
<input name="dateYear01" id="dateYear01">
<input name="calendario" class="calendario" id="datepicker01">
</div>
<div class="pickergroup">
<input name="dateDay02" id="dateDay02">/
<input name="dateMonth02" id="dateMonth02">/
<input name="dateYear02" id="dateYear02">
<input name="calendario" class="calendario" id="datepicker02">
</div>
而且日期選擇器時:
$(function() {
$(".pickergroup").find('[id^="datepicker"]').datepicker({
firstDay: 0,
monthNames: ['January', 'February', 'March',
'April', 'May', 'June',
'July', 'August', 'September',
'October', 'November', 'December'
],
dayNames: ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday'],
dayNamesShort: ['S', 'M', 'T', 'W', 'T', 'F', 'S'],
dayNamesMin: ['S', 'M', 'T', 'W', 'T', 'F', 'S'],
dateFormat: 'dd-mm-yy',
onSelect: function (selectedDate) {
var mycontainer = $(this).parent('.pickergroup');
var id = $(this).attr('name').substring(10,12);
var selectedDay = selectedDate.substring(0, 2);
var selectedMonth = selectedDate.substring(3, 5);
var selectedYear = selectedDate.substring(6);
mycontainer.find('#dateDay' + id).val(selectedDay);
mycontainer.find('#dateMonth' + id).val(selectedMonth);
mycontainer.find('#dateYear' + id).val(selectedYear);
}
});
});
如何選擇正確的父母點擊date02時?
爲什麼都有相同的ID「datepicker」? –
爲兩個datepicker使用不同的'id'。 – Azim