我有以下困难:
我正在从mysql下拉列表中获取值，我希望该信息显示在该下拉列表中。
看到这个:
<select id="location" name="location" class='form-control'>
<option value="0">Select location</option>
<?php
$query = mysql_query("select cityname from city");
while($row = mysql_fetch_assoc($query))
{
echo '<option value="'.$row['cityname'].'">'.$row['cityname']. '</option>';
}
?>
</select>
通过使用这段代码，我正在填充从数据库到下拉列表的值，但是为此，我需要刷新页面以显示值。
谢谢。
使用jQuery Ajax
yourfile.php
<select id="location" onchange="getState(this.value)" name="location" class='form-control'>
<option value="0">Select location</option>
<?php
$query = mysql_query("select * from city");
while($row = mysql_fetch_assoc($query))
{
echo '<option value="'.$row['cityid'].'">'.$row['cityname']. '</option>';
}
?>
</select>
<select id="state">
</select>
function getState(city_id)
{
var html = $.ajax({
type: "POST",
url: "path/to/ajax/my_ajax.php",
data: "city_id=" +city_id,
async: false
}).responseText;
if(html){
$("#state").html(html);
}
}
AJAX.php
$query = mysql_query("select * from state where city_id=".$_REQUEST['city_id']);
while($row = mysql_fetch_assoc($query))
{
echo '<option value="'.$row['state_id'].'">'.$row['state_name']. '</option>';
}