如何获取在php中的ajax中选择的下拉值?

时间:2016-02-10 09:11:35

标签: php jquery ajax

我对选择的ajax下拉值=在编辑或刷新时选择的问题感到困惑

  <div class="form-group" id="maincat">
                    <label for="cat_desc" class="col-sm-2 control-label">Select State</label>

                        <div class="col-sm-10">
                        <select name="city_state" id="stateDrp">  
                                <option value="">Please Select State</option>  
                        </select>
                        </div>
                </div>

 function sel_city(val){
                 $.ajax({  
                    url:"<?php echo  
                    base_url();?>tdbadmin/countriesmanagement/getstate_drpdwn",  
                    data: {id: val},  
                    type: "POST",  
                    success:function(data){  
                    $("#stateDrp").html(data);
                    //$("#stateDrp").val(<?php echo $citydata[0]->country_id;?>);   
                 }  
              }); 

        }

setTimeout(function(){
sel_city(<?php echo $citydata[0]->country_id;?>);

}, 1000);   
$("#stateDrp option[value='<?php echo     $citydata[0]->country_id;?>']").attr("selected", "selected"); 

</script>

这是我尝试的但是我无法在下拉中获得所选值,请帮助

1 个答案:

答案 0 :(得分:0)

试试这个:

$("#stateDrp option[value='<?php echo $citydata[0]->country_id;?>']").attr("selected", true);