我有脚本显示所有城市(在选择框)关于我之前选择的areaID。ajax脚本与选择框
例如 - 这我选择areaID表示2我将在第二个选择框所有瞬移到区域城市看到2
的问题是 - 当我加载页面,要证明是前所选择的城市(来自数据库)
城市选择框中的CITY选项不是指向数据库中的cityID。
我需要改变什么?
谢谢先进。
<select name='areaID' id='areaID'>
<?PHP
$query = mysql_query("SELECT * FROM `areas` ORDER BY id ASC ");
while($index = mysql_fetch_array($query))
{
$db_area_id = $index['id'];
$db_area_name = $index['name'];
if ($db_area_id == $userDetails['area'])
echo "<option value='$db_area_id' selected>$db_area_name</option>";
else
echo "<option value='$db_area_id'>$db_area_name</option>";
}
?>
</select>
<select id='cityID' name='cityID'> </select>
<script>
<?PHP if ($_POST) { ?>
$(document).ready(function(){
$('#areaID').filter(function(){
var areaID=$('#areaID').val();
var cityID=<?PHP echo $userDetails['cityID'] ?>;
$('#cityID').load('ajax/getCities.php?areaID=' + areaID+'&cityID=' + cityID);
return false;
});
});
<?PHP }else { ?>
$(function() {
function updateCitySelectBox() {
var areaID = $('#areaID').val();
$('#cityID').load('ajax/getCities.php?areaID=' + areaID);
return false;
}
updateCitySelectBox();
$('#areaID').change(updateCitySelectBox);
});
<?PHP } ?>
</script>
getCities.php:
$areaID = (int) $_GET['areaID'];
$second_option = "";
$query2 = mysql_query("SELECT * FROM `cities` WHERE area_id = $areaID ORDER BY id ASC");
while($index = mysql_fetch_array($query2))
{
$id = $index['id'];
$name = $index['name'];
$name = iconv('windows-1255', 'UTF-8', $name);
$second_option .= "<option value='$id'>$name</option>";
}
echo $second_option;
你能告诉我们getCities.php –
肯定: \t $ areaID表示=(INT)$ _GET [ 'areaID表示']; \t \t $ second_option =“”; \t \t $ query2 = mysql_query(“SELECT * FROM'cities' WHERE area_id = $ areaID ORDER BY id ASC”); \t while($ index = mysql_fetch_array($ query2)) \t { \t \t $ id = $ index ['id']; \t \t $ name = $ index ['name']; \t \t \t \t $ name = iconv('windows-1255','UTF-8',$ name); \t \t $ second_option。=“”; \t \t \t} \t \t回声$ SECOND_OPTION; – Roi
我将它添加到主帖@LiamAllan – Roi