ondown on dropdownlist

时间:2010-04-19 15:18:31

标签: jquery ajax

我的问题是继续我所要求的链接。 Load Country/State/City

我已经扩展到从db加载我的下拉列表,我只需要在我的第一个下拉列表中连接onchange方法,第二,请参阅代码。感谢任何帮助。

附加最新代码:

<select id="country"  onchange="getStateByCountryId()"></select> <br />
<select id="state"></select>  <br />


$(document).ready(function() { 
     var options = {
         type: "POST",
         url: "SearchPage.aspx/LoadCountry",
         data: "{}",
         contentType: "application/json; charset=utf-8",
         dataType: "json",

         success: function(msg) {

             var returnedArray = msg.d;
             country = $("#country"); 
              country.append('<option>Select a Country</option>'); 

             for (i = 0; i < returnedArray.length; i++) {
                  country.append("<option value='" + returnedArray[i].Id + "'>" + returnedArray[i].Name + "</option>");
             }


         }
     };
     $.ajax(options);
 });


function getStateByCountryId() {

     $("#country").change(function() 
     { 
         var _selected = $("#country").val();
         var options = 
         {
             type: "POST",
             url: "SearchPage.aspx/StateBy",
             data: "{'countryId':'" + _selected + "'}",
             contentType: "application/json; charset=utf-8",
             dataType: "json",

             success: function(msg) {
                $('#state').empty(); 
                 var returnedArray = msg.d;


                 state = $("#state");
                 for (var i = 0; i < returnedArray.length; ++i) {
                     state.append("<option value='" + returnedArray[i].Id + "'>" + returnedArray[i].Name + "</option>");
                 }
             }
         };
         $.ajax(options);
     });
 }

但不填充?我这样做的方式是你想怎么做?

感谢。

2 个答案:

答案 0 :(得分:25)

$("#state").change(function(){
    //on-change code goes in here.
    //variable "this" references the state dropdown element
});

答案 1 :(得分:3)

$(this).parent("select").attr("id");