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我試圖代表其他下拉列表的值填充下拉菜單。這裏的問題是,我正在嘗試調用select tag的onChange事件的函數未被編譯。並在此處顯示console.enter代碼中的錯誤。 HTML:Jquery/AJAX函數無法正常工作Change事件
<select name="street_id" id="street_id" onchange="get_data(this)" class="input-xlarge">
</select>
的Jquery:
<script type="text/javascript">
function get_data()
{
$('#client_list').empty()
var dropDown = document.getElementById("street_id");
var street_id = dropDown.options[dropDown.selectedIndex].value;
$.ajax({
type: "GET",
url: "http://localhost:7777//index.php/admin/get_one_street_client/" + street_id,
data: { 'street_id': street_id },
success: function(data){
// Parse the returned json data
var opts = $.parseJSON(data);
// Use jQuery's each to iterate over the opts value
$.each(opts, function(i,d) {
console.log(d.client_name);
// You will need to alter the below to get the right values from your json object. Guessing that d.id/d.modelName are columns in your carModels data
<?print "var value == '$balance_data[0]->client_id'";
//$selected = 'selected';
?>
$('#client_list').append('<option value="' + d.client_id + ' ">' + d.client_name + '</option>');
});
}
});
}
</script>
錯誤:
insert_balance:180
Uncaught ReferenceError: get_data is not definedonchange @ insert_balance:180
我已經添加在上面jQuery庫地址。
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
你在哪裏添加了你的腳本? –
這個'<?print「var value =='$ balance_data [0] - > client_id'」; // $ selected ='selected'; ?>' –
@KinshukLahiri是這是錯誤..謝謝.. – user1960257