【发布时间】:2019-03-01 20:11:11
【问题描述】:
我有一个名为 country 的表单域和一个名为 state 的域。假设我选择印度。这意味着我只想显示印度的哪些城市。假设我选择美国意味着我只想显示美国的哪些城市,从这里我写了选择查询这个和所有正确的,但我不能在状态字段中附加值..console.log(res);
这里我有正确的值,但我不能在状态字段中附加这个值
<script type="text/javascript">
function getState(country_id){
//console.log(country_id);
$.ajax({
url:'getstate.php',
type:'POST',
data : { 'country_id' : country_id},
success:function(data){
var res=jQuery.parseJSON(data);// convert the json
console.log(res);
if(res['status']==1){
var htmlString='';
$.each( res['data'], function( key, value ) {
htmlString+='<select>';
htmlString+='<option>'+value.state_name+'</option>';
htmlString+='</select>';
//console.log( key + ": " + value.name );
});
$('#state').empty().append(htmlString);
}
},
});
};
</script>
getstate.php
<?php
include('dbconfig.php');
$country_id = $_POST['country_id'];
$sql = mysql_query("SELECT * FROM state WHERE country_id='$country_id'");
$count = mysql_num_rows($sql);
$return=array();
if($count > 0){
while($row=mysql_fetch_assoc($sql)){
$data[]=$row;
}
$return=array('status'=>1,'count'=>$count,'data'=>$data);
}else{
$return=array('status'=>0,'count'=>$count,'data'=>'');
}
echo json_encode($return);
?>
<tr>
<th>COUNTRY</th>
<td>
<select name="country" id="country" onchange="getState(this.value);">
<option value="">Select Country</option>
<?php
include("dbconfig.php");
$sql = mysql_query("SELECT * FROM country");
while($row=mysql_fetch_assoc($sql)){
?>
<option value="<?php echo $row['country_id'];?>"><?php echo $row['country_name'];?></option>
<?php } ?>
</select>
<span id="country_err"></span>
</td>
</tr>
<tr>
<th>STATE</th>
<td>
<select name="state" id="state">
<option value="">Select State</option>
</select>
<span id="state_err"></span>
</td>
</tr>
【问题讨论】:
标签: javascript php jquery ajax