<form:select id="businesstype" name="businesstype" path="businesstype" >
<c:forEach items="${businesstype}" var="items">
<option value="${items.businessTypeMstrId}" >"${items.businessTypeName}" </option>
</c:forEach>
</form:select>
<select id="stateId" multiple="multiple">
<option value="">Select State</option>
</select>
<script>
$("select#businesstype").change(function(){
$.getJSON("reguser/loadStates",{countryId: 2}, function(j){
//alret(j);
var options = '';
for (var i = 0; i < j.length; i++) {
//options += '<option value="' + j[i].id + '">' + j[i].name + '</option>';
options += '<option value="' + j[i]+ '">' + j[i] + '</option>';
// alert(j[i]);
var option = document.createElement("option");
option.text = j[i];
// option.value = j[i];
var select = document.getElementById("productSelect");
select.appendChild(option);
alert(select);
}
$('stateId').empty().append('<option value=1>My option</option><option value=1>My option77</option>'); //This line is not working
});
});
</script>
嗨工作動態,我是新來的jQuery的時候我給它一個jQuery事件中我下面的代碼不能正常工作,但是當我把它裏面的JavaScript沒有任何jQuery的事件它的工作。我的完整代碼如上所述。jQuery的多選是不是平變化
$('stateId').empty().append('<option value=1>My option</option><option value=1>My option77</option>'); //This line is not working
不工作意味着改變事件沒有被解僱,是嗎? –
$('stateId')假設爲Class或ID $('#stateId')或$('。stateId') –