Question

我正在加载客户列表，我想根据选择的客户加载第二个下拉列表。出于某种原因，我的代码不起作用。这是我的代码：

<span style="display:inline-block;">
    <select name="sCustomer" id="sCustomer" onChange="findLocations(this.value)">
        <option value="0">- Select Customer -</option>
    </select>
</span>
<span style="display:inline-block;">
    <select id="sLocation" name="sLocation">
        <option value="0">- Select Location -</option>
    </select>
</span>

<script type="text/javascript">

function findLocations(custID) {
    $('#sLocation').empty();
    $('#sLocation').append("<option value='0'>- Select Location -</option>");
    $.ajax({
    type:"POST",
    url:"getLocations.php",
    contentType:"application/json; charset=utf-8",
    data: { custID : custID },
    dataType:"json",
    success: function(data){ 
        $.each(data,function(i, item){
            $('#iTest').val("bobby");
            $('#sLocation').empty();
            $('#sLocation').append("<option value='0'>- Select Location -</option>");
            $('#sLocation').append('<option value="' + data[i].id + '">' + data[i].name + '</option>');
        });
    },
    complete: function(){

    }
});
}

</script>

并且它调用的页面具有此代码

<?php 
include_once(INCLUDE_DIR.'class.error.php');
include_once(INCLUDE_DIR.'class.role.php');
include_once(INCLUDE_DIR.'class.passwd.php');
include_once(INCLUDE_DIR.'class.user.php');
include_once(INCLUDE_DIR.'class.auth.php');
include_once(INCLUDE_DIR.'class.location.php');

function getLocations() {  
    $custID = 0; 

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

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

    $locations = array();

    if (($res=db_query($sql)) && db_num_rows($res)) {
        while(list($id,$name)=db_fetch_row($res)) { 
            $columns = array (
                    'locID' => $id,
                    'locName' => $name,
            );
            $locations[] = $columns;
        }
    }
  return $locations;
}

?>

为什么这不起作用，无论如何都要测试哪个部分正在破坏？我无法回应任何事情，因为页面没有回发（因为它是ajax），我无法在ajax函数中执行javascript警报。请记住，我没有添加填充客户下拉列表的代码，因为我知道它正常工作，填充所有内容并给我正确的值。

好的，我发现这至少可以帮助我开始调试。它失败但我发现了这个

error: function (request, error) {
    console.log(arguments);
    alert(" Can't do because: " + error);
},

Answer 1

替换

data: { custID : custID },

与

data: { 'custID' : custID },

在javascript中

Answer 2

用这个替换你的php文件中的代码

if (($res=db_query($sql)) && db_num_rows($res)) {
        $str = "";
        $str .= "<option value='0'>Select Location</option>";

        while(list($id,$name)=db_fetch_row($res)) { 

              $str .= "<option value=".$id.">".$name."</option>";
        }
    }
echo $str;

现在修改你的javascript函数

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

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

}
});
}

Answer 3

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

$.ajax({
    type:"POST",
    url:"getLocations.php",
    contentType:"application/json; charset=utf-8",
    data: { custID : custID },
    dataType:"json",
    success: function(data){ 
       document.getElementById("sLocation").innerHTML = "<option value='0'>- Select Locationss -</option>"; // + data;
        },
        complete: function(){

        }
    });
}

，php页面是

 <?php 
include_once(INCLUDE_DIR.'class.error.php');
include_once(INCLUDE_DIR.'class.role.php');
include_once(INCLUDE_DIR.'class.passwd.php');
include_once(INCLUDE_DIR.'class.user.php');
include_once(INCLUDE_DIR.'class.auth.php');
include_once(INCLUDE_DIR.'class.location.php');


$custID = 0; 

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

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

$locations = array();

if (($res=db_query($sql)) && db_num_rows($res)) {
    $str = "";
    $str .= "<option value='0'>Select Location</option>";

    while(list($id,$name)=db_fetch_row($res)) { 
          $str .= "<option value=".$id.">".$name."</option>";
    }
}
echo $str;

?>

从第二个下拉列表Ajax加载下拉列表到PHP无法正常工作

3 个答案: