我有這樣的Java腳本:填入其他選擇(包括jqtransform)選擇
$("#judet div.jqTransformSelectWrapper ul li a").click(function(){
var jud= $("#judetul1").val();
$.ajax({
type: "POST",
url: "rental/cms/inc/ajax/cities.php",
data: { 'jud': jud },
success: function (msg) {
$("#oras").html(msg);
},
error: function (xhr, err) {
alert("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
alert("responseText: " + xhr.responseText);
}
});
});
和這個網站:
<div class=" h">
<span class="block">Orasul</span>
<div class="select6" id="oras">
<select name="oras1" id="oras1" onchange="zone1();sectorul();">
</select>
</div>
<div class="clear"></div>
</div>
<div class="clear"></div>
這個PHP:
public function get_oras($code3) {
echo "<option selected='selected' value='0'>Alege oras</option>";
$code='PPLA';
$code2='PPLA2';
$sql="SELECT * FROM `locatii` WHERE (`feature_code`=:code OR `feature_code`=:code2) AND `admin1_code`=:code3 ORDER BY `asciiname` ASC";
$stmt = $this->dbh->prepare($sql);
$stmt->bindParam(':code', $code, PDO::PARAM_STR, 30);
$stmt->bindParam(':code2', $code2, PDO::PARAM_STR, 30);
$stmt->bindParam(':code3', $code3, PDO::PARAM_INT);
$stmt->execute();
foreach ($stmt->fetchAll(PDO::FETCH_ASSOC) as $result)
{
$oras[]="<option value='".$result['geonameid']."'>".$result['asciiname']."</option>";
}
return $oras;
}
的HTML JUDETUL :
<div class=" h">
<span class="block">Judetul</span>
<div class="select6" id="judet">
<?php $judetul=$db->get_judet(); ?>
<select name="judetul1" id="judetul1" >
<option selected="selected">---</option>
<?php foreach ($judetul as $val=>$k) { ?>
<option value="<?php echo $val; ?>"><?php echo $k; ?></option>
<?php } ?>
</select>
</div>
<div class="clear"></div>
</div>
<div class="clear"></div>
問題是: 我需要在更改judetul1時填充oras1。 當我更改judetul1選項時,選擇oras1不會從數據庫獲取值。
我該如何解決?
你的HTML中的「judetul1」在哪裏?你已經知道,如果JS,PHP或兩者都失敗了? – kero
如果我刪除jqtransform腳本,js和php工作正常! –