Question

我试图从ajax调用加载下拉列表。这是我的代码：

<select name="customer" onChange="testAJAX(this.value)">
    <option value="0" selected >Select Customer</option>
    <?php
        $services = db_query('SELECT cust_id,cust_name FROM '.CUST_TABLE.' ORDER BY cust_name');

        while (list($cust_id,$name) = db_fetch_row($services)){
            $selected =($info['customer']==$cust_id)?'SELECTED':'';?>
                <option value="<?=$cust_id?>"<?=$selected?>><?=$name?></option>
                    <?php
                     }?>
</select>

Javascript功能

function testAJAX(custID) {
$('#sLocation').empty();

var myURL = "getLocations.php?custID=" + custID;

$.ajax({
    type:"POST",
    url: myURL,
    contentType:"application/json; charset=utf-8",
    data: { custID : custID },
    dataType:"json",
    success: function(data){ 
        document.getElementById("sLocation").innerHTML = data; 
    },
    error: function (xhr, textStatus, errorThrown) {
            document.getElementById("sLocation").innerHTML = xhr.responseText; 
    },
    complete: function(){

    }
});

}

和它调用的页面

<?php 

$custID = 0;
$mysqli = new mysqli("localhost", "James", "mypassword", "OSTickets"); 

if (isset($_REQUEST['custID'])) {
    $custID = $_REQUEST['custID'];
}

$sql = 'SELECT l.location_id, l.site_name FROM ost_customer_location' 
. ' AS l JOIN ost_customer'
. ' AS c ON c.cust_id=l.cust_id'
. ' WHERE c.cust_id='.$custID;

$locations = array();

$str = "";
$str .= "<option value='0'>Select Location</option>";
if ($result = $mysqli->query($sql)) {
    while ($row = mysqli_fetch_array($result)){ 
        $str .= "<option value=".$row["0"].">".$row["1"]."</option>";
    }
}
echo $str;
?>

我可以让它工作，但只能在ajax的错误功能而不是成功但是当我提醒xhr.responseText时，它没有显示任何错误，只是选择的html代码是：

<option value='0'>Select Location</option>
<option value='252'>Location 1</option>
<option value='345'>Location 2</option>

我如何弄清楚它投掷的错误？

将错误功能更改为

error: function(jqXHR, textStatus, errorThrown){ console.log(jqXHR); }

这是来自控制台的响应。

对象{readyState：4，responseText：＆＃34; Duke Rd - Primary＆＃34;，状态：200，statusText：＆＃34; OK＆＃34;}

Ajax请求永远不会成功，但没有错误

0 个答案: