Javascript Ajax选择并自动加载数据
我的表格:Javascript Ajax选择并自动加载数据,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我的表格: <form> <select id="f_name"><?php echo $fname ?></select> <select id="l_name></select> </form> Myfullname.php: $db = mysqli_connect("localhost", "root", "", "test"); if(isset($_POST['name'])){ $f_
<form>
<select id="f_name"><?php echo $fname ?></select>
<select id="l_name></select>
</form>
Myfullname.php
:
$db = mysqli_connect("localhost", "root", "", "test");
if(isset($_POST['name'])){
$f_name = $_POST['name'];
$sql = mysqli_query($db, "SELECT last_name FROM fullname WHERE first_name = '$f_name'");
$res = mysqli_fetch_array($sql);
echo $res;
}
当我选择名字时,姓氏不会出现。您的表单格式不正确:
<form>
<select id="f_name"><?php echo $fname ?></select>
<select id="l_name></select>
</form>
$(文档).ready(函数(e){
$(“#f#u name”).change(函数(){
var fname=$('this').val();
$.ajax({
类型:'POST',
数据:{name:fname},
url:“fullname.php”,
成功:函数(数据){
对于数据中的(i){
$(“#l#u name”).html(“+data[i]+”);
}
}
});
});
});
thank you do this work“$('#l#u name').html(''+data+'')”如果输入了我的标记,我仍然一无所获:(你能成功记录数据吗?并告诉我们它是什么吗
<form>
<select id="f_name"><?php echo $fname ?></select>
<select id="l_name></select>
</form>
<form>
<select id="f_name">
<option><?php echo $fname ?></option>
</select>
<select id="l_name></select>
</form>
$(document).ready(function(e){
$("#f_name").change(function(){
var fname = $('this').val();
$.ajax({
type :'POST',
data :{name:fname},
url :"fullname.php",
success : function(data){
$('#l_name').html('<option>'+data+'</option>');
}
});
});
});
$(document).ready(function(e){
$("#f_name").change(function(){
var fname = $('this').val();
$.ajax({
type :'POST',
data :{name:fname},
url :"fullname.php",
success : function(data){
for (i in data) {
$("#l_name").html("<option>'+data[i]+'</option>");
}
}
});
});
});